Question

cos(θ+15)−tan(θ−15)=(4cos2θ)÷(1+2sin2θ)

Answer 1

@ikram002p pleas help me

Answer 2

@ikram002p please

Answer 3

@ganeshie8 @ikram002p @zepdrix sorry for extra tagging

Answer 4

cos(theta - 15)) or there is division sign in between??

Answer 5

Sorry, + is there..

Answer 6

no its cos (blah) - tan (blah) = blah

Answer 7

yea its ok c:

Answer 8

@ikram002p is having her BOYFRIEND!!! WHAT DOES THIS MEAN!??!?!?!?!?!??!

Answer 9

bf = breakfast lolololol

Answer 10

SHE IS HAVING HER BOYFRIEND FOR BREAKFAST, OH MER GUSH :OOOOOOOOOOOOOOOOO

Answer 11

ahhahahaahhacant argue with that but any pls

Answer 12

pls help me on the original question

Answer 13

I just want to know that if we can use the fraction values of cos(15) or sin(15) here???

Answer 14

idk if u can

Answer 15

The 15 degree angle is on the LHS but not on the RHS.
Expand the LHS and get rid of the 15 degree angle by substituting (45-30) in its place since the trig values for 45 and 30 degrees are known.

Answer 16

show me how that would wrk in the fraction bcos it doesnt seem t work

Answer 17

\(
\sin(15) = \sin(45-30) = \sin(45)\cos(30) - \cos(45)\sin(30) = \\
\sqrt{2}/2 * \sqrt{3}/2 - \sqrt{2}/2 * 1/2= \sqrt{2}/4 * (\sqrt{3} - 1)
\)

Answer 18

@aum It is just like using fraction values of sin(15) and cos(15)..

Answer 19

@aum what about theat

Answer 20

\[\sin(15^{\circ}) = \frac{\sqrt{3}-1}{2 \sqrt{2}}\]
\[\cos(15^{\circ}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}}\]
And hence:

\[\tan(15^{\circ}) = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\]

Answer 21

My calculation is still going lengthy and it is not reaching anywhere significant.. :(

Answer 22

One way I started by using tan(x) = sin(x)/cos(x)..

But that will lead to:

\[\frac{\sqrt{3} + 2\cos(2 \theta) - 8\sin(\theta - 15)}{8\cos(\theta - 15)}\]

Answer 23

May be I have chosen wrong path to go with.. I try it again.. :)

Answer 24

It may be worthwhile first to make sure the identity is true for say theta = 0.

Answer 25

And I don't think the given identity is true for theta = 0.

Answer 26

Yeah, it is not true for given value : \(\theta\)..

Answer 27

\(\theta\) being 0..

Answer 28

Instead of an identity, you can treat this as an equation instead and solve for theta using a graphing calculator.

Answer 29

http://www.wolframalpha.com/input/?i=cos%28%CE%B8%2B15%29%E2%88%92tan%28%CE%B8%E2%88%9215%29%3D%284cos2%CE%B8%29%C3%B7%281%2B2sin2%CE%B8%29