Trigonometric Functions - Math

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Simplify the function below:

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We need to use the following formulas:

a)

and

b)

We can simplify as follows:

Which of the following is not in the range of the function $y = $\sin{x}$$ ?

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The range of the function $y= \sin x$ is all numbers between and (the sine wave never goes above or below this).

Of the choices given, $\frac{3}{2}$ is greater than and thus not in this range.

Simplify the following trionometric function:

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

Replace the trigonometric functions with these values:

$$\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}$

Change a angle to radians.

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

Therefore, to solve:

Simplify the following trigonometric function in fraction form:

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

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

Replacing these values, we get:

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

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

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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}$}=5$

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

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$x=\frac{\pi}{3}$\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}$}{2}$

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

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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$

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

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

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

If , give the value of .

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We know that .

Simplify:

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We know that .

Then we can write:

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

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

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$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)$

Simplify the following expression:

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

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

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

Simplify the trigonometric expression.

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

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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$

Give the value of :

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

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$sin($\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}$=0.5$

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

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

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$tan\ x=\frac{1}{4}$\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}$$

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

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

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

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

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

Simplify the following expression:

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We need to use the following identitities:

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

Now substitute them into the expression:

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

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

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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

Which of the following is equivalent to

$\tan\theta\csc\theta$ ?

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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)}$= \sec(\theta)$