# Given sinx=4/5, cosy=7/25 find the following.

• February 10th 2009, 12:59 AM
Random-Hero-
Given sinx=4/5, cosy=7/25 find the following.
Angles x and y are located in the first quadrant such that sinx=4/5, and cosy = 7/25.

a) Determine an exact value of cosx.

b) Determine an exact value of siny.

c) Determine an exact value for sin(x+y)

I'm really at a loss here, do I just sub those in to some formula or something? Can anyone help me out? Or get me pointed in the right direction? thanks in advance!!
• February 10th 2009, 01:31 AM
Air
Quote:

Originally Posted by Random-Hero-
Angles x and y are located in the first quadrant such that sinx=4/5, and cosy = 7/25.

a) Determine an exact value of cosx.

This is a $3-4-5$ triangle. If $\sin x = \frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}=\fra c{4}{5}$ then what is $\cos x = \frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}} = ...$.

Quote:

Originally Posted by Random-Hero-
b) Determine an exact value of siny.

This is a $7-24-25$ triangle. If $\cos y = \frac{\mathrm{Adjacent}}{\mathrm{Hypotenuse}} = \frac{7}{25}$ then what is $\sin y = \frac{\mathrm{Opposite}}{\mathrm{Hypotenuse}}= ...$.

Quote:

Originally Posted by Random-Hero-
c) Determine an exact value for sin(x+y)

I'm really at a loss here, do I just sub those in to some formula or something? Can anyone help me out? Or get me pointed in the right direction? thanks in advance!!

$\sin(x+y) = \sin x\cos y + \sin y \cos x$ thus once you have values of $\sin x, \cos y, \sin y, \cos x$, you can solve $\sin(x+y)$.
• February 10th 2009, 01:38 AM
Trigonometry
Hello Random-Hero
Quote:

Originally Posted by Random-Hero-
Angles x and y are located in the first quadrant such that sinx=4/5, and cosy = 7/25.

a) Determine an exact value of cosx.

b) Determine an exact value of siny.

c) Determine an exact value for sin(x+y)

I'm really at a loss here, do I just sub those in to some formula or something? Can anyone help me out? Or get me pointed in the right direction? thanks in advance!!

The formulae you need here are:

$\cos^2 A + \sin^2\ A = 1$ and $\sin(A + B) = \sin A\cos B + \cos A \sin B$

So in (a), use $\cos^2 x = 1 - \sin^2x$, and remember when you take the square root to find $\cos x$, that, since the angle $x$ is in the first quadrant $\cos x > 0$, so you don't have to worry about the negative square root.

(b) Use $\sin^2y = 1 - cos^2y$. And, again, since $y$ is in the first quadrant, $\sin y > 0$.

(c) simply use the formula for $\sin(A+B)$ and your answers to (a) and (b).

Can you complete it now?