1. ## help with question

Hi,

Can anyone help me with part c please

2. ## Re: help with question

Originally Posted by dabatman
Hi,

Can anyone help me with part c please
$\therefore cos x = \frac {\sqrt 5}{3} and tan x = \frac {2}{\sqrt 5}\\ \\ and sin y = \frac {\sqrt 7}{4} and tan y = \frac {\sqrt 7}{3} \\ \\ tan (x+y) = \frac {tanx + tan y}{1 - tan x \times tan y}$

Then just substitute the values of tan x and of tan y indicated above.

3. ## Re: help with question

$\noindent You will have to make use of the \tan (\cdot) expansion. \\ \tan (x+y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} \\ Multiplying the numerator and denominator by \cos x \cos y, the fraction simplifies to \\ \frac{\tan x + \tan y}{1 - \tan x \tan y} = \frac{\sin x \cos y + \sin y \cos x }{\cos x \cos y - \sin x \sin y} . \\ You can then use this new expression to evaluate the original expression. (Note that the fraction can also be expressed as \frac{\sin (x+y)}{\cos (x+y)} (which you can double-check by expanding the \sin and \cos).$

4. ## Re: help with question

thank you so much

5. ## Re: help with question

Also,

Can someone help me with this question, i solved it but my answer was incorrect

6. ## Re: help with question

$sin 3\theta = sin (\theta + 2\theta) = sin\theta cos2\theta + cos\theta sin 2\theta \\ \\ = sin\theta (1-2sin^2 \theta) + cos\theta (2sin\theta cos\theta) = sin\theta - 2 sin^3 \theta + 2sin\theta (1-sin^2 \theta)\\ \\ = 3sin\theta - 4sin^3 \theta$

7. ## Re: help with question

Originally Posted by Drongoski
$sin 3\theta = sin (\theta + 2\theta) = sin\theta cos2\theta + cos\theta sin 2\theta \\ \\ = sin\theta (1-2sin^2 \theta) + cos\theta (2sin\theta cos\theta) = sin\theta - 2 in^3 \theta + 2sin\theta (1-sin^2 \theta)\\ \\ = 3sin\theta - 4sin^3 \theta$
thankyou

8. ## Re: help with question

Originally Posted by dabatman
thankyou
I know that this question was from a while ago, but the answer from the book says its "3sintheater x(times) cos2theater - sin3theater"

9. ## Re: help with question

anyone know why?

10. ## Re: help with question

Do you mean?:

$3sin\theta \times cos^2 \theta - sin^3 \theta\\ \\ = 3sin\theta (1-sin^2 \theta) - sin^3 \theta \\ \\ =3sin\theta - 3 sin^3 \theta - sin^3 \theta\\ \\ = 3sin\theta - 4sin^3 \theta$

Well this is the "unfinished" version of sin3@ to be expressed in terms of sin@ only. But looking at the question again, it only asked for an expression. The 2 are equivalent.

By the way: it is theta and not theater.

11. ## Re: help with question

Thanks, and oops spelling ain't my forte, lol

There are currently 1 users browsing this thread. (0 members and 1 guests)

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•