trig identities?

Question

Can't figure out how to do this, help please!

1.


sin( (3π/2) + x ) + sin ( (3π/2) - x ) = - 2 cos x


2.

..........θ............ csc θ - cotθ
sin² ----- = --------------------
... . . ..2...............2cscθ


one more, how exactly do you find the exact value of cot 75˚ using half-angle formulas?

Answer 1

1) sin( (3π/2) + x ) + sin ( (3π/2) - x ) = -2cos(x)

First, we choose the more complex side, which would be the left hand side.

LHS = sin( (3π/2) + x ) + sin ( (3π/2) - x )

Use the sine addition(and subtraction) identities:
sin(a + b) = sin(a)cos(b) + sin(b)cos(a)
sin(a - b) = sin(a)cos(b) - sin(b)cos(a).

LHS = [ sin(3π/2)cos(x) + sin(x)cos(3π/2) ] +
[ sin(3π/2)cos(x) - sin(x)cos(3π/2) ]

We know that sin(3π/2) = -1 and cos(3π/2) = 0, so

LHS = (-1)cos(x) + sin(x)(0) + (-1)cos(x) - sin(x)(0)
LHS = -cos(x) + (-1)cos(x)
LHS = -cos(x) - cos(x)
LHS = -2cos(x) = RHS

2) sin²(θ/2) = [csc(θ) - cot(θ)] / [2csc(θ)]

Choose the more complex side, which would be the right hand side.

RHS = [ csc(θ) - cot(θ) ] / [2csc(θ)]

Convert everything to sines and cosines.

RHS = [ 1/sin(θ) - cos(θ)/sin(θ) ] / [ 2(1/sin(θ)) ]

Multiply top and bottom by sin(θ), to get rid of complex fractions.

RHS = [ 1 - cos(θ) ] / 2

Here's the part which gets tricky; use the double angle identity for cosine. As a reminder,

cos(2x) = cos²(x) - sin²(x). BUT, if x = θ/2, then
cos(θ) = cos²(θ/2) - sin²(θ/2)

RHS = [ 1 - [ cos²(θ/2) - sin²(θ/2) ] ] / 2

Distribute the minus,

RHS = [ 1 - cos²(θ/2) + sin²(θ/2) ] / 2

Use the identity 1 - cos²(θ/2) = sin²(θ/2)

RHS = [ sin²(θ/2) + sin²(θ/2) ] / 2

RHS = [ 2sin²(θ/2) ] / 2

Cancel the 2,

RHS = sin²(θ/2) = LHS

3) cot (75˚)

By definition, this is equal to

cos(75˚) / sin(75˚)

Let's use the half-angle formula to solve for each. A reminder that
cos²(x) = (1/2)(1 + cos(2x))
sin²(x) = (1/2)(1 - cos(2x))
cot²(x) = (1 + cos(2x)) / (1 - cos(2x))


cot²(75˚) = (1 + cos(150˚)) / (1 - cos(150˚))

But cos(150˚) = -√(3)/2, so

cot²(75˚) = (1 + √(3)/2 ) / ( 1 - √(3)/2 )

Getting rid of the complex fraction, multiply top & bottom by 2.

cot²(75˚) = (2 + √(3)) / (2 - √(3))

Rationalize by multiplying top and bottom by the bottom's conjugate.

cot²(75˚) = (2 + √(3))² / (2² - 3)

cot²(75˚) = (4 + 2√(3) + 3) / (4 - 3)

cot²(75˚) = 7 + 2√(3)

Take the square root of both sides; normally we would have two solutions (plus and minus), but since we know which quadrant 75˚ lies in, we're going to take the negative solution (in the second quadrant, cot is negative).

cot(75˚) = -√( 7 + 2√(3) )

Answer 2

1. Remember that sine is just cosine shifted over by pi/2 (or 3pi/2 depending on how you look at it) and this one should become easy.

2. I never write csc, cot, or sec. Rewrite these in terms of sin and cos at it gets easy (multiply by sin on top and bottom of the RHS)

3. You should be able to figure out exact values of cos 150 and sin 150 by looking at the unit circle. Use the half angle formulas on these to get the sin and cos of 75 and simply divide them.

Answer 3

Question 1
sin(3π/2 + x) = sin3π/2.cosx + cos3π/2.sinx = - cosx
sin(3π/2 - x) = sin3π/2.cosx - cos3π/2.sinx = - cos x
sum = - 2 cos x

Question 2
RHS = [1/sin Ø - cos Ø / sin Ø] / 2 / sin Ø
RHS = [1 - cos Ø] / 2
RHS = [1 - (1 - 2sin ² Ø/2)] / 2
RHS = sin ² Ø/2 = LHS as required

Question 3
cos(45 + 30) / sin(45 + 30)
Top line = N = (1/√2).(√3/2) - (1/√2).(1/2)
Bottom line = D = (1/√2).(√3/2) + (1/√2).(1/2)
N/D = (√3 - 1) / (√3 + 1)
N/D = (√3 - 1)² /(√3 + 1).(√3 - 1)
N/D = (√3 - 1)² / 2
N/D= (1/2).(3 - 2√3 + 1)
N/D = (1/2).(4 - 2√3)
N/D = 2 - √3

Answer 4

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Answer 5

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