Trigonometric Identities - Pre-Calculus

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Question

Find $\sin\bigg(\frac{\pi }{3} + \frac{\pi }{6}\bigg)$ using the sum identity.

Answer

Using the sum formula for sine,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where,

$a=\frac{\pi}{3}$ , $b=\frac{\pi}{6}$

yeilds:

$\sin \bigg(\frac{\pi}{3}+\sin\frac{\pi}{6}\bigg)=\sin\bigg(\frac{\pi }{3}\bigg)\cdot\cos\bigg(\frac{\pi }{6}\bigg) +\cos \bigg(\frac{\pi}{3}\bigg)\cdot\sin\bigg(\frac{\pi }{6}\bigg)$

$=\frac{\sqrt{3}}2{}\cdot\frac{\sqrt{3}}2{}+\frac{1}{2}\cdot\frac{1}{2}$

$=\frac{3}{4}+\frac{1}{4}=1$ .

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Question

Calculate $\sin(75)$ .

Answer

Notice that $\sin(75)$ is equivalent to $\sin(30+45)$ . With this conversion, the sum formula can be applied using,

$\sin(a+b)=\sin(a)\cos(b)+\cos(a)\sin(b)$

where

, .

Therefore the result is as follows:

$\sin(a+b)=\sin(30)\cdot \cos(45)+\cos(30)\cdot\sin(45)$

$=\frac{1}{2}\cdot \frac{\sqrt{2}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{\sqrt{2}}2{}$

$=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}=\frac{\sqrt2 +\sqrt6}{4}$ .

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Question

Evaluate the exact value of:

Answer

In order to solve , two special angles will need to be used to solve for the exact values.

The angles chosen are and degrees, since:

Write the formula for the cosine additive identity.

Substitute the known variables.

$cos(30)cos(45)-sin(30)sin(45)=\left(\frac{\sqrt3}{2}\right)\left(\frac{\sqrt2}{2}\right)-\left(\frac{1}{2}\right)\left(\frac{\sqrt2}{2}\right)$

$=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}=\frac{\sqrt6-\sqrt2}{4}$

$2cos(75)=2\left(\frac{\sqrt6-\sqrt2}{4}\right)=\frac{\sqrt6-\sqrt2}{2}$

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\cos (\alpha + \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ . So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the cosine sum formula, we then see:

$\ \cos (\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta\ \* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} - \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} - \sqrt {15}}{20}$ .

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\cos (\alpha - \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the cosine difference formula, we see:
$\ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta\ \* = \frac {2 \sqrt {6}}{5} \cdot \frac {1}{4} + \frac {1}{5} \cdot \frac {\sqrt {15}}{4} = \frac {2 \sqrt{6} + \sqrt {15}}{20}$

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha + \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine sum formula, we see:

$\ \sin (\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \ \* = \frac {1}{5} \cdot \frac {1}{4} + \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 + 6 \sqrt {10}}{20}$

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\sin (\alpha - \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\cos \alpha = \frac {2\sqrt{6}}{5}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\sin \beta = \frac {\sqrt{15}}{4}$ .

Using the sine difference formula, we see:

$\ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \ \* = \frac {1}{5} \cdot \frac {1}{4} - \frac {2 \sqrt {6}}{5} \cdot \frac {\sqrt {15}}{4} = \frac {1 - 6 \sqrt {10}}{20}$

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan (\alpha + \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So
$\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\ \tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \ \ \* = \frac{\frac{\sqrt{6}}{12}+\sqrt{15}}{1-\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}+12\sqrt{15}}{12}}{\frac{12-3\sqrt{10}}{12}} \ \ \* = \frac{\sqrt{6}+12\sqrt{15}}{12-3\sqrt{10}} \cdot \frac{12+3\sqrt{10}}{12+3\sqrt{10}} \ \ \* = \frac{12\sqrt{6}+6\sqrt{15}+144\sqrt{15}+180\sqrt{6}}{144-90} \ \* = \frac{192\sqrt{6}+150\sqrt{15}}{54} = \frac{32\sqrt{6}+25\sqrt{15}}{9}$

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Question

In the problem below, $\sin \alpha = \frac {1}{5}, 0 < \alpha < \frac {\pi}{2}$ and $\cos \beta = \frac {1}{4}, 0 < \beta< \frac {\pi}{2}$ .

Find

$\tan(\alpha - \beta)$ .

Answer

Since $\sin \alpha = \frac {1}{5}$ and $\alpha$ is in quadrant I, we can say that and and therefore:

$x = \sqrt {r^2 - y^2} = \sqrt {25-1} = \sqrt {24} = 2 \sqrt {6}$ .

So $\tan \alpha = \frac {1}{2\sqrt{6}} = \frac {\sqrt{6}}{12}$ .

Since $\cos \beta = \frac {1}{4}$ and $\beta$ is in quadrant I, we can say that and and therefore:

$y = \sqrt {r^2 - x^2} = \sqrt {16-1} = \sqrt {15}$ .

So $\tan \beta = \sqrt{15}$ .

Using the tangent sum formula, we see:

$\ \tan(\alpha-\beta) = \frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta} \ \ \* = \frac{\frac{\sqrt{6}}{12}-\sqrt{15}}{1+\frac{\sqrt{6}}{12} \cdot \frac{\sqrt{15}}{1}}}} = \frac{\frac{\sqrt{6}-12\sqrt{15}}{12}}{\frac{12+3\sqrt{10}}{12}} \ \ \* = \frac{\sqrt{6}-12\sqrt{15}}{12+3\sqrt{10}} \cdot \frac{12-3\sqrt{10}}{12-3\sqrt{10}} \ \ \* = \frac{12\sqrt{6}-6\sqrt{15}-144\sqrt{15}+180\sqrt{6}}{144-90} \ \ \* = \frac{192\sqrt{6}-150\sqrt{15}}{54} = \frac{32\sqrt{6}-25\sqrt{15}}{9}$

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Question

Find the value of .

Answer

To solve , we will need to use both the sum and difference identities for cosine.

Write the formula for these identities.

$cos(A\pm B)= cos(A)cos(B)\pm sin(A)sin(B)$

To solve for and , find two special angles whose difference and sum equals to the angle 15 and 75, respectively. The two special angles are 45 and 30.

Substitute the special angles in the formula.

$cos(45\pm 30)= cos(45)cos(30)\pm sin(45)sin(30)$

Evaluate both conditions.

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)+\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$=\frac{\sqrt2}{2}\left(\frac{\sqrt3}{2}\right)-\frac{\sqrt2}{2}\left(\frac{1}{2}\right)=\frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Solve for .

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\left[\frac{\sqrt6}{4}-\frac{\sqrt2}{4}\right]$

$=\frac{\sqrt6}{4}+\frac{\sqrt2}{4}-\frac{\sqrt6}{4}+\frac{\sqrt2}{4}$

$= \frac{2\sqrt2}{4}$

$= \frac{\sqrt2}{2}$

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Question

Given that $\tan x = 2$ and $\tan y = 3$ , find $\tan (x+y)$ .

Answer

Jump straight to the tangent sum formula:

$\ \tan (x+y) = \frac {\tan x + \tan y}{1-\tan x \tan y}$

From here plug in the given values and simplify.

$\ \ = \frac {2+3}{1-2\cdot3} \ \= \frac {5}{-5} \ \= -1$

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Question

Find the exact value for: $4\sin (15^{\circ})$

Answer

In order to solve this question, it is necessary to know the sine difference identity.

The values of and must be a special angle, and their difference must be 15 degrees.

A possibility of their values that match the criteria are:

Substitute the values into the formula and solve.

$sin(15)=(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})-(\frac{\sqrt2}{2})(\frac{1}{2})= \frac{\sqrt6}{4}-\frac{\sqrt2}{4}$

Evaluate .

$=4(\frac{\sqrt6}{4}-\frac{\sqrt2}{4})= \sqrt6-\sqrt2$

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Question

Find the exact value of: $\sin(105^{\circ})$

Answer

In order to find the exact value of , the sum identity of sine must be used. Write the formula.

The only possibilites of and are 45 and 60 degrees interchangably. Substitute these values into the equation and evaluate.

$sin(105)=(\frac{\sqrt2}{2})(\frac{1}{2})+(\frac{\sqrt2}{2})(\frac{\sqrt3}{2})=\frac{\sqrt2+\sqrt6}{4}$

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Question

Which of the following expressions best represents $\tan\left ( 60^{\circ}\right )$ ?

Answer

Write the identity for $tan(2\theta)$ .

$tan(2x)=\frac{2tan(x)}{1-tan^2(x)}$

Set the value of the angle equal to .

Substitute the value of into the identity.

$\frac{2tan(30)}{1-tan^2(30)}$

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Question

Evaluate

$\sin \left(\frac{5 \pi}{12}\right)$ .

Answer

$\frac{5 \pi }{12}$ is equivalent to $\frac{2 \pi }{12 } + \frac{3 \pi }{12 }$ or more simplified $\frac{\pi}{6 } + \frac{\pi }{4}$ .

We can use the sum identity to evaluate this sine:

$\sin \left(\frac{\pi }{6} + \frac{\pi }{4 } \right) = \sin\left (\frac{\pi }{6} \right) \cos \left(\frac { \pi }{4} \right) + \cos \left(\frac{\pi }{6}\right) \sin \left( \frac{ \pi }{4} \right)$

From the unit circle, we can determine these measures:

$\frac{1}{2} \cdot \frac{1}{\sqrt 2 } + \frac{ \sqrt 3 }{2} \cdot \frac{1}{ \sqrt 2 } = \frac{1}{2 \sqrt 2 }+ \frac{\sqrt 3 }{2 \sqrt 2 } = \frac{ 1 + \sqrt 3 }{ 2 \sqrt 2 }$

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Question

Evaluate

$\sin(285 ^ o)$ .

Answer

The angle or .

Using the first one:

$\sin(60^o + 225^o ) = \sin (60^o) \cos (225^o) + \cos (60^o) \cos(225^o )$

We can find these values in the unit circle:

$\frac{\sqrt 3 }{2 } \cdot \frac{-1}{\sqrt 2 } + \frac{1}{2} \cdot \frac{-1}{\sqrt 2 } = \frac{-\sqrt3 }{2 \sqrt 2 } + \frac{-1}{2\sqrt2} = \frac{-\sqrt 3 - 1 }{2 \sqrt 2 }$

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Question

Use trigonometric identities to solve the following equation for $\theta$ :

$tan^2\theta + sec^2\theta = 1$

Answer

Use the trigonometric identities to switch sec into terms of tan:

$tan^2\theta + sec^2\theta = tan^2\theta + (tan^2 \theta + 1) = 2tan^2\theta + 1 = 1$

hence,

$2tan^2\theta = 0$

So we have $\tan \theta = 0$ , making $\theta= n \pi$

Therefore the solution is $\theta = n \pi$ for n being any integer.

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Question

Which of the following is not a solution to $\small 2sin^2(x)-3sin(x)=-1$ for $\small 0\leq x<2\pi$

Answer

We begin by setting the right side of the equation equal to 0.

$\small 2sin^2(x)-3sin(x)+1=0$

The equation might be easier to factor using the following substitution.

$\small w=sin(x)$

This gives the following

$\small 2w^2-3w+1=0$

This can be factored as follows

$\small \small (2w-1)(w-1)=0$

Therefore

$\small w=\frac{1}{2};or;w=1$

Replacing our substitution therefore gives

$\small sin(x)=\frac{1}{2};or;sin(x)=1$

Within our designated domain, we get three answers between our two equations.

$\small sin(x)=\frac{1}{2}$ when $\small x=\frac{\pi}{6};or; \frac{5\pi}{6}$

$\small sin(x)=1$ when $\small x=\frac{\pi}{2}$

Therefore, the only choice that isn't correct is $\small \frac{11\pi}{6}$

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Question

Find one possible value of .

Answer

Begin by isolating the tangent side of the equation:

Next, take the inverse tangent of both sides:

$5x=tan^{-1} (-72)$

Divide by five to get the final answer:

$x=\frac{tan^{-1} (-72)}{5}=-17.84^{\circ}$

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Question

Use trigonometric identities to solve for the angle value.

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13$

Answer

There are two ways to solve this problem. The first involves two trigonometric identities:

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\ \implies\tan^2\theta+1=13\implies\sec^2\theta=13\ \implies\sec\theta=\sqrt13\implies\cos\theta=\frac{\sqrt13}{13}\ \implies \theta=\cos^{-1}\frac{\sqrt13}{13}}\approx73.898}$

The second method allows us to only use the first trigonometric identity:

$\tan^2\theta\sin^2\theta+\tan^2\theta\cos^2\theta+1=13\ \implies\tan^2\theta(\sin^2\theta+\cos^2\theta)+1=13\ \implies\tan^2\theta+1=13\implies\tan^2\theta=12\ \implies\tan\theta=2\sqrt3\implies\theta=\tan^{-1}2\sqrt3\approx73.898}$

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