Question #86f6d

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Question #86f6d

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Answer 1 · 2016-07-12T15:00:11

Draw a couple of triangles and make use of the sum formula for sine to get $sin ({cos}^{- 1} (- \frac{33}{65}) + {tan}^{- 1} (\frac{13}{35})) = \frac{1531}{65 \sqrt{1394}}$ $sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1531/(65sqrt(1394))$ .

Explanation:

These questions require us to remember what sine and cosine and tangent really are: relationships among legs of a right triangle. So, how would you find the sine of two different angles added together ? Draw two different right triangles using the information given in the problem and use the sum formula for sine.

If we let $x = {cos}^{- 1} (- \frac{33}{65})$ $x=cos^(-1)(-33/65)$ and $y = {tan}^{- 1} (\frac{13}{35})$ $y=tan^(-1)(13/35)$ , then we really have $sin (x + y)$ $sin(x+y)$ . And if you recall, $sin (x + y) = sin x cos y + sin y cos x$ $sin(x+y)=sinxcosy+sinycosx$ . So all we really need to do is find what $sin x$ $sinx$ , $sin y$ $siny$ , $cos x$ $cosx$ , and $cos y$ $cosy$ are equal to.

We already know what $cos x$ $cosx$ is equal to. If $x = {cos}^{- 1} (- \frac{33}{65})$ $x=cos^(-1)(-33/65)$ , then $cos x = - \frac{33}{65}$ $cosx=-33/65$ , using the inverse relationship between cosine and inverse cosine. But how do we find $sin x$ $sinx$ ? Simple - draw a triangle! Remember that $cos x = \frac{adjacent}{hypotenuse}$ $cosx="adjacent"/"hypotenuse"$ , so $cos x = - \frac{33}{65}$ $cosx=-33/65$ means we have a triangle with an adjacent side of $- 33$ $-33$ and a hypotenuse of $65$ $65$ . Using the Pythagorean Theorem, we can solve for the length of the other side:
${hypotenuse}^{2} = {adjacent}^{2} + {opposite}^{2}$ $"hypotenuse"^2="adjacent"^2+"opposite"^2$
$65^{2} = {(- 33)}^{2} + {opposite}^{2}$ $65^2=(-33)^2+"opposite"^2$
$3136 = {opposite}^{2}$ $3136="opposite"^2$
$56 = opposite$ $56="opposite"$

Now that we know all the side lengths, we can construct the triangle:
enter image source here
From the diagram, it is clear that $sin x$ $sinx$ , which is opposite divided by hypotenuse, is $\frac{56}{65}$ $56/65$ . We have $sin x$ $sinx$ and $cos x$ $cosx$ , now we only need $sin y$ $siny$ and $cos y$ $cosy$ . Finding them will be almost the same thing.

If $y = {tan}^{- 1} (\frac{13}{35})$ $y=tan^(-1)(13/35)$ , then $tan y = \frac{13}{35}$ $tany=13/35$ . Since tangent is defined as opposite over adjacent, we have a triangle with an opposite side of $13$ $13$ and an adjacent side of $35$ $35$ . The Pythagorean Theorem says:
${hypotenuse}^{2} = {adjacent}^{2} + {opposite}^{2}$ $"hypotenuse"^2="adjacent"^2+"opposite"^2$

Solving for the hypotenuse, we get:
${hypotenuse}^{2} = {(35)}^{2} + {(13)}^{2}$ $"hypotenuse"^2=(35)^2+(13)^2$
${hypotenuse}^{2} = 1394$ $"hypotenuse"^2=1394$
$hypotenuse = \sqrt{1394}$ $"hypotenuse"=sqrt(1394)$

Here's this triangle:
enter image source here

We can see that $sin y = \frac{13}{\sqrt{1394}}$ $siny=13/sqrt(1394)$ and $cos y = \frac{35}{\sqrt{1394}}$ $cosy=35/sqrt(1394)$ .

Let's remind ourselves of what we found so far:
$sin x = \frac{56}{65}$ $sinx=56/65$
$cos x = - \frac{33}{65}$ $cosx=-33/65$
$sin y = \frac{13}{\sqrt{1394}}$ $siny=13/sqrt(1394)$
$cos y = \frac{35}{\sqrt{1394}}$ $cosy=35/sqrt(1394)$

Using the sum formula for sine, we can obtain our result:
$sin (x + y) = sin x cos y + sin y cos x$ $sin(x+y)=sinxcosy+sinycosx$
$\to sin ({cos}^{- 1} (- \frac{33}{65}) + {tan}^{- 1} (\frac{13}{35})) = (\frac{56}{65}) (\frac{35}{\sqrt{1394}}) + (\frac{13}{\sqrt{1394}}) (- \frac{33}{65})$ $->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=(56/65)(35/sqrt(1394))+(13/sqrt(1394))(-33/65)$
$\to sin ({cos}^{- 1} (- \frac{33}{65}) + {tan}^{- 1} (\frac{13}{35})) = \frac{1960}{65 \sqrt{1394}} - \frac{429}{65 \sqrt{1394}}$ $->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1960/(65sqrt(1394))-429/(65sqrt(1394))$
$\to sin ({cos}^{- 1} (- \frac{33}{65}) + {tan}^{- 1} (\frac{13}{35})) = \frac{1960}{65 \sqrt{1394}} - \frac{429}{65 \sqrt{1394}}$ $->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1960/(65sqrt(1394))-429/(65sqrt(1394))$
$\to sin ({cos}^{- 1} (- \frac{33}{65}) + {tan}^{- 1} (\frac{13}{35})) = \frac{1531}{65 \sqrt{1394}}$ $->sin(cos^(-1)(-33/65)+tan^(-1)(13/35))=1531/(65sqrt(1394))$

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Question #86f6d

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