Simplifying Trig functions...urgent help needed

[Samara]

these are my last three problems...i am stuck on these problems...please help

 

[Loren]

In the first one, your second line is wrong. Take one step at a time.

$\displaystyle \cos^2B-\sin^2B=2\cos^2B-1$
$\displaystyle \cos^2B-(1-\cos^2B)=2\cos^2B-1$

Now, remove the parenthesis and proceed.

 

[Samara]

Am I on the right track on he other two problems?

 

[galactus]

For the last one, are you sure that's the correct equation?.

This this an identity to prove or an equation to solve?. If it's an identity to prove, in general $\displaystyle csc(x)-cos(x)\neq{sin(x)tan(x)}$

If it's an equation:

$\displaystyle csc(x)-cos(x)=sin(x)tan(x)$

$\displaystyle \frac{1}{sin(x)}-cos(x)=\frac{sin^{2}(x)}{cos(x)}$

Multiply through by cos(x):

$\displaystyle sin^{2}(x)=cot(x)-cos^{2}(x)$

$\displaystyle sin^{2}(x)+cos^{2}(x)=cot(x)$

Now, can you finish?. Notice the left side, what that is?.

 

[o_O]

For the second one, try converting the csc[sup:2uo4fvvs]2[/sup:2uo4fvvs]t - 1 to cot(t) first and then convert everything in terms of sin and cos:

$\displaystyle cost(csc^{2}t - 1) = cost \cdot cot^{2}t = cost \cdot \frac{cos^{2}t}{sin^{2}t} = \frac{cos^{3}t}{sin^{2}t}$

Hmm. That's annoying. We have a power of 2 on the bottom and a power of 3 on the top. What can we multiply both top and bottom by :wink: ?

 

[galactus]

Is this the 3rd one?.

$\displaystyle \frac{cot^{3}(t)}{csc(t)}=cos(t)(csc^{2}(t)-1)$

If so,

$\displaystyle \frac{cot^{3}(t)}{csc(t)}=\frac{\frac{cos^{3}(t)}{sin^{3}(t)}}{\frac{1}{sin(t)}}=\frac{cos^{3}(t)}{sin^{3}(t)}\cdot\frac{sin(t)}{1}=\frac{cos^{3}(t)}{sin^{2}(t)}=\frac{cos^{2}(t)}{sin^{2}(t)}\cdot\frac{cos(t)}{1}=cot^{2}(t)cos(t)$$\displaystyle =cos(t)(csc^{2}(t)-1)$