# Thread: A few problems...Compound & double-angle trig. formulae

1. ## A few problems...Compound & double-angle trig. formulae

Im stuck on a few questions...

Any ideas?

1) Solve the following equation for θ in the interval 0 ≤ θ ≤ 2π.

sin( 2 θ) +5/9 cos( θ) = 0

2) Solve
4 cos( 2 θ) = 1+tan( θ)

for θ in the interval −π/2 < θ < π/2.

3) Which of the following is equal to

(sin θ + cos θ)^2 + (sin θ − cos θ)^2 for all θ?

A) 2 cos(2 θ)
B) 2
C) 2 sin^2θ
D) cos(2 θ) − sin(2 θ)
E) 2 sin(2 θ)

2. 1) sin( 2*beta ) + 5/9 * cos( beta ) = 0

sin( 2*beta ) = 2*sin( beta )*cos( beta )

sin( 2*beta ) + 5/9 * cos( beta ) = 2*sin( beta )*cos( beta ) + 5/9 * cos( beta ) = 0

cos( beta )*[ 2*sin( beta ) + 5/9 ] = 0

a) cos( beta ) = 0 => beta = pi/2 or beta = 3*pi/2

b) 2*sin( beta ) + 5/9 = 0

sin( beta ) = -5/18 => beta = arcsin( -5/18 ) ..... use calculator

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2) it is more complicated.... probably will solve it later

// i ' ve tried to solve it but it is complcated so the answer is very long and i didnt finished it, probably there is some faster way , some kind of trick...

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3)
the right answer is B) 2

( sin( beta ) + cos( beta ) )^2 + ( sin( beta ) - cos( beta ) )^2 =
= [ sin^2( beta ) + 2*sin( beta )*cos( beta ) + cos^2( beta ) ] + [ sin^2( beta ) - 2*sin( beta )*cos( beta ) + cos^2( beta ) ] =
= 2*( sin^2( beta ) + cos^2( beta ) ) = 2*1 = 2

3. Originally Posted by josipive
1) sin( 2*beta ) + 5/9 * cos( beta ) = 0

sin( 2*beta ) = 2*sin( beta )*cos( beta )

sin( 2*beta ) + 5/9 * cos( beta ) = 2*sin( beta )*cos( beta ) + 5/9 * cos( beta ) = 0

cos( beta )*[ 2*sin( beta ) + 5/9 ] = 0

a) cos( beta ) = 0 => beta = pi/2 or beta = 3*pi/2

b) 2*sin( beta ) + 5/9 = 0

sin( beta ) = -5/18 beta = arcsin( -5/18 ) ..... use calculator
Shouldnt there be 4 answers...or I could be way off...

4. Originally Posted by Maccabhoy
Shouldnt there be 4 answers...or I could be way off...
2 different answers for cos( beta ) = 0
and 2 two different answers for arcsin( -5/18 )

for arcsin( -5/18 ) calculator will give you answer between -pi/2 and pi/2
but you have one more answer and that is arcsin( -5/18 ) - pi/2 ( I think )

also look at the first post to see answer for 3)

// not arcsin( -5/18 ) - pi/2 , but - pi - arcsin( -5/18 )

5. Thanks josipive.

The explanations are great!

6. Hello, Maccabhoy!

Here's the last one . . .

3) Which of the following is equal to . $(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2$ .for all $\theta$ ?

. . $(A)\;2\cos(2\theta) \qquad(B)\;2 \qquad
Did you even try to multiply it out ??

$(\sin\theta + \cos\theta)^2 + (\sin\theta -\cos\theta)^2$

. . $= \;\sin^2\!\theta + 2\sin\theta\cos\theta + \cos^2\!\theta + \sin^2\!\theta - 2\sin\theta\cos\theta + \cos^2\!\theta$

. . $= \;2\sin^2\!\theta + 2\cos^2\!\theta \;=\;2\underbrace{(\sin^2\!\theta + \cos^2\!\theta)}_{\text{This is 1}} \;=\;2(1) \;=\;2$

7. Originally Posted by josipive
2 different answers for cos( beta ) = 0
and 2 two different answers for arcsin( -5/18 )

for arcsin( -5/18 ) calculator will give you answer between -pi/2 and pi/2
but you have one more answer and that is arcsin( -5/18 ) - pi/2 ( I think )

also look at the first post to see answer for 3)

// not arcsin( -5/18 ) - pi/2 , but - pi - arcsin( -5/18 )
Hmm... arcsin (-5/18) and - pi - arcsin( -5/18 ) are giving me wring answers.

I am getting -0.28 and -2.86 respectively for these, but the quiz says they are wrong answers...and yes my calculator is in rad...

8. anyone know how figure this one out?

2) Solve
4 cos( 2 θ) = 1+tan( θ)

for θ in the interval −π/2 < θ < π/2.

9. Originally Posted by Maccabhoy
Hmm... arcsin (-5/18) and - pi - arcsin( -5/18 ) are giving me wring answers.

I am getting -0.28 and -2.86 respectively for these, but the quiz says they are wrong answers...and yes my calculator is in rad...
dont know... on my calculator everything works fine

1.angle ) arcsin( -5/18 )

2.angle ) -pi - arcsin( -5/18 ) // try -pi + arcsin( 5/18 ) that is the same