Trig - Finding Angles

**Nick87** · June 25th 2008, 04:01 PM

Having trouble with my Trig these days. I was wondering if you could help me out?

Q. If Angle B is acute and TanB = $\frac 3{4}$

Find sinB, cosB, sin2B and tan2B

- - -

Also, could anyone help me with another solving for 0 ≤ x ≤ 360 question?

Q. $4 - sinx = 4cos^2x$

Methods would be great. Thanks! Always appreciative of response

**badgerigar** · June 25th 2008, 04:14 PM

Q. If Angle B is acute and TanB =

Find sinB, cosB, sin2B and tan2B

using $\tan = \frac{opposite}{adjacent}$ we can draw a right angle triangle like this:

can you continue?

For the last 2 you may like to use these:
$\sin(2x) = 2\sin(x)\cos(x)$
$\cos(2x) = \cos^2(x)-sin^2(x)$
There are other formulas you might know but these are the ones I have memorised.

Also, could anyone help me with another solving for 0 ≤ x ≤ 360 question?

Q.

The first step uses $\sin^2(x)+\cos^2(x)=1$

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