Sum and differences of angles?

**eh501** · November 3rd 2008, 02:40 PM

I'm having a little trouble, here's the question:

If sin $x= 7/25$ and cos $y= 3/5$ , and $x$ and $y$ are positive acute angles, find

a. $sin (x+y)$
b. $cos (x-y)$
c. $tan (x+y)$

**Jameson** · November 3rd 2008, 03:17 PM

First question - do you know the sum and difference formulas?

**eh501** · November 3rd 2008, 03:48 PM

Yeah.

**Soroban** · November 3rd 2008, 05:45 PM

Hello, eh501!

Okay, you know the three formulas:

. . $\begin{array}{ccc}\sin(x+y) &=&\sin x\cos y + \cos x\sin y \\ \\[-3mm] \cos(x+y) &=&\cos x\cos y - \sin x\sin y \\ \\[-3mm] \tan(x+y) &=&\dfrac{\tan x + \tan y}{1 - \tan x\tan y} \end{array}$

If $\sin x= \tfrac{7}{25}\text{ and }\cos y= \tfrac{3}{5}$ , and $x$ and $y$ are positive acute angles, find

. . $a)\;\;\sin (x+y)\qquad b)\;\;\cos (x-y) \qquad c)\;\;\tan (x+y)$

To use those formulas, we need more trig values.

We can use: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1$

Since $\sin x = \tfrac{7}{25}\!:\;\;(\tfrac{7}{25})^2 + \cos^2\!x \:=\:1 \quad\Rightarrow\quad\cos^2\!x \:=\:1-\tfrac{49}{625}$

. . . . $\cos^2\!x\:=\:\tfrac{576}{625} \quad\Rightarrow\quad \cos x \:=\:\tfrac{24}{25}$

Then: . $\tan x \;=\;\frac{\sin x}{\cos x} \;=\;\frac{\frac{7}{25}}{\frac{24}{25}} \;=\;\tfrac{7}{24}$

. . Hence, we have: . $\begin{Bmatrix} \sin x &=& \frac{7}{25} \\ \\[-4mm] \cos x &=& \frac{24}{25} \\ \\[-4mm] \tan x &=& \frac{7}{24} \end{Bmatrix}$

Use the same procedure on $\cos y = \tfrac{3}{5}$ to find $\sin y\text{ and }\tan y$

. . then you can use those formulas . . .

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