# trig function questions (1 Viewer)

#### Schniz

##### Member
1)
Simplify in terms of the indicated angle:
1-2sin^2 3A; in terms of 6A

2)
Simplify:
cosA(cos4A+2sin^2 2A)

3)
Prove:
(cos@+sin@)/(cos@-sin@) = sec2@+tan2@

4)
Write down expressions for tan3@ and cos3@ in terms of the angle @. (argh what kind of question i sthis??? :S)

5)
Evaluate the integral from pi/4 to 0 of (cos^2 @ -1)

6)
Write down a formula for sin3@ in terms of the angle @

thanks... any help is muchly appreciated

#### SoulSearcher

##### Active Member
i) 1-2sin23A = cos6A
ii) cosA(cos4A+2sin22A) =
cosA(1-2sin22A+2sin22A)
= cosA
iii) LHS = (cos@+sin@)/(cos@-sin@)
= (cos@+sin@)/(cos@-sin@) * (cos@+sin@)/(cos@+sin@)
= (cos2@+sin2@+2sin@cos@)/{cos2@-sin2@)
= (1+sin2@)/cos2@
= sec2@+tan2@
= RHS
iv) tan3@ = tan(2@+@)
= (tan2@+tan@)/(1-tan@tan2@)
= {[2tan@/(1-tan2@)] + tan@}/(1-tan@[2tan@/(1-tan2@)])
= (3tan@-tan3@/1-tan2@)/(1-3tan2@/1-tan2@)
= 3tan@-tan3@/1-3tan2@
cos3@ = cos(2@+@)
= cos2@cos@ - sin2@sin@
= 2cos3@ - cos@ - (2cos@ - 2cos3@)
= 4cos3@ - 3cos@

Last edited:

#### A l

##### Member
5) 1 if it is 0 to pi/4, -1 if it is actually pi/4 to 0

#### Schniz

##### Member
thanks... im not sure if i understand how you did the first one... im fairly dumb...

#### ianc

##### physics is phun!
Question 5: (i did not get AI's answer)

cos<sup>2</sup>@-1 = 1/2 + 1/2(cos2@)-1
= -1/2 + 1/2(cos2@)

Hence the integral of cos<sup>2</sup>@-1 =
integral -1/2 + 1/2(cos2@)
= [-@/2 + 1/4sin2@] + c

then substituting in the limits of pi/4 and 0:
= -pi/8 + 1/4
= -0.143 (3 decimal places)

------

Question 6: (just using the double angle formula and the identity cos<sup>2</sup>@+sin<sup>2</sup>@=1)

sin3@
= sin@cos2@ + cos@sin2@
= sin@(cos<sup>2</sup>@ - sin<sup>2</sup>@) + cos@(2sin@cos@)
= sin@cos<sup>2</sup>@ - sin@<sup>3</sup>@ + 2cos<sup>2</sup>@sin@
= sin@(1 - sin<sup>2</sup>@) - sin@<sup>3</sup>@ + 2sin@(1 - sin<sup>2</sup>@)
= 3sin@ - 4sin<sup>3</sup>@

Last edited:

#### ianc

##### physics is phun!
for the first question, you use the double angle formula for cos2A

fully worked out:
1 - 2sin<sup>2</sup>3A
= 1 - sin<sup>2</sup>3A - sin<sup>2</sup>3A
= 1 - (1-cos<sup>2</sup>3A) - sin<sup>2</sup>3A
= cos<sup>2</sup>3A - sin<sup>2</sup>3A
= cos[2(3A)]
=cos6A

hope this helps!

#### SoulSearcher

##### Active Member
vi) sin 3@ = sin (2@+@)
= sin2@cos@ + cos2@sin@
= 2sin@cos2@ + (1-2sin2@)sin@
= 2sin@(1-sin2@) + (1-2sin2@)sin@
= 3sin@ - 4sin3@

thanks everyone