Compound Angle Formulas

**skeske1234** · March 28th 2009, 04:06 PM

Can someone describe to me clearly, how you would solve these questions:

1. Angles x and y are located in the first quadrant such that sinx=4/5 and cosy=7/25.
a) determine an exact value for cos x
b) determine an exact value for sin y
c) determine an exact value for sin(x+y)

2. angle x lies in the third quadrant, and tanx=7/24
a) determine an exact value for cos2x
b) determine an exact value for sin2x

for 1 a and b I got the answers, but for 1 c, I am not sure how to approach it

**EitanG** · March 28th 2009, 04:27 PM

1) for sin(x) = 4/5
you can construct the triangle with opposite length 4 and hypotenuse 5. you can find angle x using sin^-1. then find the adjacent side and find cos(x).
use similar methods for part b. use double angle forumla for part c.

2) construct the triangle like you did for part 1, then solve

**skeske1234** · March 28th 2009, 04:48 PM

for part c of question 1, if I get to
sin(4/5)cos(7/25)+cos(3/5)sin(24/25)
what do I do next?

**Grandad** · March 29th 2009, 08:41 AM

Hello skeske

Originally Posted by skeske1234

Can someone describe to me clearly, how you would solve these questions:

1. Angles x and y are located in the first quadrant such that sinx=4/5 and cosy=7/25.
a) determine an exact value for cos x
b) determine an exact value for sin y
c) determine an exact value for sin(x+y)

This is what you need to know to solve a question like this:

If an angle is in the first quadrant, then its sine and cosine are both positive.
The relation between sine and cosine is $\sin^2x+\cos^2x=1$

So: $\sin x =\frac45 \Rightarrow \left(\frac45\right)^2 +\cos^2x = 1$

$\Rightarrow \cos^2x = 1 - \frac{16}{25}=\frac{9}{25}$

$\Rightarrow \cos x = \frac{3}{5}$ , since $\cos x > 0$

In the same way, if $\cos y = \frac{7}{25}, \sin y = \frac{24}{25}$

Originally Posted by skeske1234

for part c of question 1, if I get to
sin(4/5)cos(7/25)+cos(3/5)sin(24/25)
what do I do next?

You're getting things a bit confused here.

$\sin(x+y) =\sin x\cos y + \cos x \sin y = \frac{4}{5}\cdot \frac{7}{25} + \frac{3}{5}\cdot \frac{24}{25} = \frac{28}{125}+ \frac{72}{125}= \frac{100}{125} = \frac{4}{5}$

2. angle x lies in the third quadrant, and tanx=7/24
a) determine an exact value for cos2x
b) determine an exact value for sin2x

If angle $x$ lies in the third quadrant, then $\sin x$ and $\cos x$ are both negative.

If you know the value of $\tan x$ , then use:

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

and then use $\cos^2x = \frac{1}{\sec^2 x}$

Then, $\cos 2x = 2\cos^2x - 1$

Finally, for $\sin 2x$ , the quickest way is to use $\sin^22x = 1 - \cos^22x$

Note that $\sin 2x$ will be positive, since $\sin x$ and $\cos x$ are both negative and $\sin 2x = 2 \sin x \cos x$

Can you complete this? If not, let's see your working, and we'll finish it for you.

Grandad

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