Simplifying Trigonometric Functions - Trigonometry

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Question

Simplify the following trionometric function:

$\frac{4sin^2(60^{\circ})-tan(45^{\circ})}{3sin(45^{\circ})}$

Answer

To solve the problem, you need to know the following information:

$sin(60^{\circ}) = \frac{\sqrt{3}}{2}$

$tan(45^{\circ})=1$

$sin(45^{\circ})=\frac{\sqrt{2}}{2}$

Replace the trigonometric functions with these values:

$\frac{4sin^2(60^{\circ})-tan(45^{\circ})}{3sin(45^{\circ})}$

$\frac{4(\frac{\sqrt{3}}{2})^2-(1)}{3(\frac{\sqrt{2}}{2})}$

$\frac{4(\frac{3}{4})-(1)}{(\frac{3\sqrt{2}}{2})}$

$\frac{2}{(\frac{3\sqrt{2}}{2})}$

$\frac{4}{3\sqrt{2}} = \frac{4\cdot \sqrt{2}}{3\sqrt{2}\cdot \sqrt{2}} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}$

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Question

Change a $-240^{\circ}$ angle to radians.

Answer

In order to change an angle into radians, you must multiply the angle by $\frac{\pi }{180^{\circ}}$ .

Therefore, to solve:

$-240^{\circ} \cdot \frac{\pi }{180^{\circ}} = \frac{-4\pi }{3}$

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Question

Simplify the following trigonometric function in fraction form:

$cos^{2}(45^{\circ})-sin(30^{\circ})$

Answer

To determine the value of the expression, you must know the following trigonometric values:

$cos(45^{\circ}) = \frac{\sqrt{2}}{2}$

$sin (30^{\circ})=\frac{1}{2}$

Replacing these values, we get:

$cos^{2}(45^{\circ})-sin(30^{\circ})$

$(\frac{\sqrt{2}}{2})^{2}-\frac{1}{2}$

$\frac{2}{4}-\frac{1}{2}=\frac{1}{2}-\frac{1}{2}=0$

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Question

If $cos\ x=\frac{1}{3}$ and $sin\ x eq 0$ , give the value of $\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}$ .

Answer

Based on the double angle formula we have, $sin\ 2x=2(sin\ x)(cos\ x)$ .

$\frac{sin\ x+sin\ 2x}{sin\ x-sin\ 2x}=\frac{sin\ x+2(sin\ x)(cos\ x)}{sin\ x-2(sin\ x)(cos\ x)}$

$=\frac{sinx(1+2cos\ x )}{sinx(1-2cos\ x )}=\frac{1+2cos\ x}{1-2cos\ x}$

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

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Question

If $x=\frac{\pi}{3}$ , give the value of $\frac{sin\ x}{sin\ x+cos\ 2x}$ .

Answer

$x=\frac{\pi}{3}\Rightarrow sin\ x=sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$

Now we can write:

$x=\frac{\pi}{3}\Rightarrow 2x=2(\frac{\pi}{3})=\frac{2\pi}{3}\Rightarrow cos\ 2x=cos(\frac{2\pi}{3})$

$=cos(\pi-\frac{\pi}{3})$

$=-cos (\frac{\pi}{3})$

$=-\frac{1}{2}$

Now we can substitute the values:

$\frac{sin\ x}{sin\ x+cos\ 2x}=\frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}-\frac{1}{2}}=\frac{\sqrt{3}}{\sqrt{3}-1}=\frac{\sqrt{3}}{\sqrt{3}-1}\times \frac{{\sqrt{3}+1}}{{\sqrt{3}+1}}=\frac{3+\sqrt{3}}{2}$

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Question

If $cot\ 34^{\circ}=1.5$ , give the value of $\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}$ .

Answer

$326^{\circ}=360^{\circ}-34^{\circ}\Rightarrow sin\ 326^{\circ}=sin(360^{\circ}-34^{\circ})\Rightarrow sin\ 326^{\circ}=-sin\ 34^{\circ}$

$56^{\circ}=90^{\circ}-34^{\circ}\Rightarrow sin\ 56^{\circ}=sin(90^{\circ}-34^{\circ})\Rightarrow sin\ 56^{\circ}=cos\ 34^{\circ}$

$304^{\circ}=270^{\circ}+34^{\circ}\Rightarrow cos\ 304^{\circ}=cos(270^{\circ}+34^{\circ})\Rightarrow cos\ 304^{\circ}=sin\ 34^{\circ}$

Now we can simplify the expression as follows:

$\frac{2sin\ 326^{\circ}+3sin\ 56^{\circ}}{cos\ 304^{\circ}}=\frac{-2sin\ 34^{\circ}+3cos\ 34^{\circ}}{sin\ 34^{\circ}}$

$=-2+3cot\ 34^{\circ}=-2+3\times 1.5=-2+4.5=2.5$

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Question

If $sin\ x\times cos\ x=\frac{1}{2}$ , what is the value of $(sin\ x+cos\ x)^2$ ?

Answer

$(sin\ x+cos\ x)^2=sin^2x+cos^2x+2sin\x cos\ x$

$=1+2(sin\ x)(cos\ x)=1+2(\frac{1}{2})=1+1=2$

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Question

If $sin^2x-cos^2x=\frac{1}{2}$ , give the value of .

Answer

We know that .

$=(cos^2x-sin^2x)\times 1=-(sin^2x-cos^2x)=-\frac{1}{2}$

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Question

Simplify:

Answer

We know that .

Then we can write:

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

$=(cos^2x-sin^2x)(\frac{cos^2x+sin^2x}{cos^2x})+tan^2x$

$\Rightarrow A=\frac{(cos^2x-sin^2x)}{cos^2x}+tan^2x$

$=\frac{cos^2x}{cos^2x}-\frac{sin^2x}{cos^2x}+tan^2x$

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Question

$tan (x)sin(x) + sec(x)(cos x)^{2} =$

Answer

$tan (x)sin(x) + sec(x)(cos x)^{2} =$

$\frac{sin (x)}{cos (x)}* sin(x) + \frac{1}{cos (x)}*(cos x)^{2} =$

$\frac{(sin x)^{2} + (cos x)^{2}}{cos (x)} = \frac{1}{cos (x)} = sec (x)$

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Question

Simplify the following expression:

$A=cos(\frac{3\pi}{2}+x)+sin (\pi+x)+cos(\frac{\pi}{2}+x)+sin(\pi-x)$

Answer

We need to use the following identities:

$cos(\frac{3\pi}{2}+x)=sin\ x$

$sin (\pi+x)=-sin\ x$

$cos(\frac{\pi}{2}+x)=-sin\ x$

$sin(\pi-x)=sin\ x$

Use these to simplify the expression as follows:

$A=cos(\frac{3\pi}{2}+x)+sin (\pi+x)+cos(\frac{\pi}{2}+x)+sin(\pi-x)$

$\Rightarrow A=sin\ x-sin\ x-sin\ x+sin\ x = 0$

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Question

Give the value of :

$A=\frac{2sin(\frac{\pi}{2})+4cos(\frac{\pi}{2})}{4sin(\frac{\pi}{2})-cos(\frac{\pi}{2})}$

Answer

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

$cos(\frac{\pi}{2})=0$

Plug these values in:

$A=\frac{2sin(\frac{\pi}{2})+4cos(\frac{\pi}{2})}{4sin(\frac{\pi}{2})-cos(\frac{\pi}{2})}=\frac{2\times 1+4\times 0}{4\times 1-0}=\frac{2}{4}=\frac{1}{2}=0.5$

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Question

If $tan\ x = \frac{1}{4}$ , solve for

$D=\frac{sin\ x}{sin\ x-cos\ x}+\frac{sin\ x+cos\ x}{cos\ x}$

Answer

$tan\ x=\frac{1}{4}\Rightarrow \frac{sin\ x}{cos\ x}=\frac{1}{4}\Rightarrow cos\ x=4sin\ x$

Substitute $cos\ x=4sin \ x$ into the expression:

$D=\frac{sin\ x}{sin\ x-cos\ x}+\frac{sin\ x+cos\ x}{cos\ x}=\frac{sin\ x}{sin\ x-4sin\ x}+\frac{sin\ x+4sin\ x}{4sin\ x}$

$\Rightarrow D=\frac{sin\ x}{-3sin\ x}+\frac{5sin\ x}{4sin\ x}$

$\Rightarrow D=-\frac{1}{3}+\frac{5}{4}=\frac{-4+15}{12}=\frac{11}{12}$

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Question

If $cot\ x= 3$ , give the value of :

$D=\frac{cos^2x}{(sin\ x)(cos\ x)}-\frac{sin\ x+cos\ x}{sin\ x}$

Answer

$cot\ x=3\Rightarrow \frac{cos\ x}{sin\ x}=3\Rightarrow cos\ x= 3sin\ x$

Now substitute $cos\ x=3sin\ x$ into the expression:

$D=\frac{cos^2x}{(sin\ x)(cos\ x)}-\frac{sin\ x+cos\ x}{sin\ x}=\frac{(3sin\ x)^2}{(sin\ x)(3sin\ x)}-\frac{sin\ x+3sin\ x}{sin\ x}$

$\Rightarrow D=\frac{9sin^2x}{3sin^2x}-\frac{4sin\ x}{sin\ x}$

$\Rightarrow D=\frac{9}{3}-4=3-4=-1$

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Question

Simplify the following expression:

$A=\frac{sin^2x}{1+tan^2x}-\frac{cos^2x}{1+cot^2x}+1$

Answer

We need to use the following identitities:

$\frac{1}{cos ^2x}=1+tan^2x$

$\frac{1}{sin^2x}=1+cot^2x$

Now substitute them into the expression:

$A=\frac{sin^2x}{1+tan^2x}-\frac{cos^2x}{1+cot^2x}+1=\frac{sin^2x}{\frac{1}{cos^2x}}-\frac{cos^2x}{\frac{1}{sin^2x}}+1$

$\Rightarrow A=(sin^2x)(cos^2x)-(sin^2x)(cos^2x)+1=0+1=1$

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Question

Which of the following is equal to $\sin x\tan x$ ?

Answer

Break $\tan x$ apart: $\tan x =\frac{\sin x }{\cos x}$ .

This means that $\sin x\tan x=\sin x *\frac{\sin x}{\cos x}$ or $\frac{\sin^2 x}{\cos x}$

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Question

Which of the following is equivalent to

$\tan\theta\csc\theta$ ?

Answer

In order to evaluate this expression, rewrite the trigonometric identity in terms of sines and cosines. The tangent is equal to the sine over the cosine and the cosecant is the reciprocal of the sine; thus, we can write the following:

$\tan(\theta)\csc(\theta)= \frac{\sin(\theta)}{\cos(\theta)}\times\frac{1}{\sin(\theta)}$

Now, can simplify. Notice that the sine terms cancel each other out.

$\frac{\sin(\theta)}{\cos(\theta)}\times\frac{1}{\sin(\theta)}= \frac{1}{\cos(\theta)}$

Remember, that the reciprocal of the cosine is the secant.

$\frac{1}{\cos(\theta)}= \sec(\theta)$

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Question

Simplify: $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$

Answer

Rewrite $\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)}$ in terms of sines and cosines.

$\frac{sin(\theta)}{tan(\theta)}-\frac{csc(\theta)}{sec(\theta)} = \frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}$

Simplify the complex fractions.

$\frac{sin(\theta)}{\frac{sin(\theta)}{cos(\theta)}}-\frac{\frac{1}{sin(\theta)}}{\frac{1}{cos(\theta)}}=sin(\theta)\cdot \frac{cos(\theta)}{sin(\theta)}-\frac{1}{sin(\theta)}\cdot \frac{cos(\theta)}{1}$

$=cos(\theta)-\frac{cos(\theta)}{sin(\theta)}= cos(\theta)-cot(\theta)$

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Question

Simplify the following expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

Answer

We will first invoke the appropriate ratio for cotangent, and then use pythagorean identities to simplify the expression:

$sec^2xcot^2x \ - sec^2x + tan^2x$

$= \frac{1}{cos^2x} * \frac{cos^2}{sin^2x} - (sec^2 x - tan^2x){}$

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

$= csc^2x - 1{}$

since

$sec^2x - tan^2x = 1; \ csc\ x = \frac{1}{sin \ x}{}$

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Question

Simplify the trigonometric expression.

$\frac{\tan(\theta)\csc(\theta)}{\sec(\theta)}$

Answer

Using basic trigonometric identities, we can simplify the problem to

$\frac{\frac{\sin(\theta)}{\cos(\theta)}\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}$ .

We can cancel the sine in the numerator and the one over cosine cancels on top and bottom, leaving us with 1.

$\frac{\frac{\sin(\theta)}{\cos(\theta)}\frac{1}{\sin(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{\frac{1}{\cos(\theta)}}{\frac{1}{\cos(\theta)}}=\frac{1}{\cos(\theta)}\cdot \frac{\cos(\theta)}{1}=1$

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