Can someone please help me solve the following trig. identity?

Question

Solve 0
1.a. sinø- tanø = 0
b. 2sin2 ø= sq. root2
c. sinø= 1 + cosø

Answer 1

I'll get you started on #1

sin(theta) - sin(theta)/cos(theta) =0

sin(theta)cos(theta) / cost(theta)
- sin(theta)/cos(theta) =0

(sin(theta)cos(theta) - sin(theta)) / cos(theta)=0

Therefore sin(theta)cos(theta) - sin(theta)=0

sin(theta) (cos(theta) - 1) =0

Now solve sin(theta)=0 and cos(theta)=1

Answer 2

1a. sinθ - tanθ = 0

sinθ - (sinθ/cosθ) = 0
(sinθ/cosθ)(cosθ - 1) = 0

sinθ = 0 => θ = k*180 for k being an integer
OR
cosθ - 1 = 0 => cosθ = 1 => θ = k*360 for k being an integer

So to keep the answer between 0 and 360 degrees, we only need to consider solutions for when k = 0, 1, 2 (possibly). Note that plugging in k > 2 would result in an answer beyond 360 for sure.

θ = 0, 180, 360 are the only answers that lie in your interval.



1b. 2sin2θ = √2

You need to get θ by itself.

sin2θ = (√2)/2

2θ = 45 + k*360 for k being an integer
OR
2θ = 135 + k*360 for k being an integer

So our answer becomes

θ = 45/2 + k*180 for k being an integer
OR
θ = 135/2 + k*180 for k being an integer

To keep the angle between 0 and 360 degrees, we just need to consider k = 0, 1

θ = 45/2, 135/2, 405/2, 495/2 are the angles the lie in your interval.



1c. sinθ = 1 + cosθ

Square both sides.
(sinθ)^2 = (1 + cosθ)^2

Expand the right side.
(sinθ)^2 = 1 + 2cosθ + (cosθ)^2

Recall the famous identity: (sinθ)^2 + (cosθ)^2 = 1
(sinθ)^2 = (sinθ)^2 + (cosθ)^2 + 2cosθ + (cosθ)^2

Subtract (sinθ)^2 from both sides.
0 = 2(cosθ)^2 + 2cosθ

Divide 2 from both sides.
0 = (cosθ)^2 + cosθ

Factor out cosθ from the right side.
0 = cosθ(cosθ + 1)

cosθ = 0 => θ = 90 + k*180 where k is an integer
OR
cosθ = -1 => θ = 180 + k*360 where k is an integer

To keep θ in the proper interval, we only really need to consider k = 0, 1

θ = 90, 180 are the only angles that lie in your interval.

Answer 3

1)
a) sin(t) - tan(t) = 0

Change the left hand side to sines and cosines.

sin(t) - sin(t)/cos(t) = 0

Put under a common denominator.

[sin(t)cos(t) - sin(t)]/[cos(t)] = 0

Factor sin(t) out of the numerator.

sin(t) [cos(t) - 1] / cos(t) = 0

Factor the fraction of 1/cos(t).

[sin(t)/cos(t)] [cos(t) - 1] = 0
[tan(t)] [cos(t) - 1] = 0

Now, equate each of those to 0 and solve.

tan(t) = 0, cos(t) - 1 = 0.

tan(t) = 0 implies t = {0, pi}
cos(t) - 1 = 0 implies cos(t) = 1, which means t = 0.

Therefore, our solutions are t = {0, pi, 2pi}, or, in degrees,
t = {0, 180, 360}

2. 2sin2(t) = sqrt(2)

Divide both sides by 2,

sin(2t) = sqrt(2)/2

Sine is equal to sqrt(2)/2 at two places: pi/4, 3pi/4. However, we must double our solutions because we have 2t. Our solutions will be:

2t = {pi/4, 3pi/4, pi/4 + 2pi, 3pi/4 + 2pi}
2t = {pi/4, 3pi/4, 9pi/4, 11pi/4}

Now, divide each solution by 2, which is the same as multiplying them by 1/2.

t = {pi/8, 3pi/8, 9pi/8, 11pi/8}
Converted to degrees,
t = {22.5, 67.5, 202.5, 247.5}

3. sin(t) = 1 + cos(t)

To solve this, let's move the cos(t) to the left hand side.

sin(t) - cos(t) = 1

Square both sides, to obtain

sin^2(t) - 2sin(t)cos(t) + cos^2(t) = 1

sin^2(t) + cos^2(t) - 2sin(t)cos(t) = 1

Note the identity sine squared x plus cos squared x = 1.

1 - 2sin(t)cos(t) = 1
-2sin(t)cos(t) = 0
2sin(t)cos(t) = 0

Note that 2sin(t)cos(t) = sin(2t) by the double angle identity, so

sin(2t) = 0, and it follows that (note: we have to double our solutions because we have 2t)

2t = {0, pi, 2pi, 3pi}. Dividing both sides (and therefore each term) by 2, we get

t = {0, pi/2, pi, 3pi/2}

Because squaring an equation may add unnecessary solutions, we have to TEST each value.

sin(t) = 1 + cos(t).

Test t = 0: sin(0) = 0, 1 + cos(0) = 1 + 1 = 2.
Reject t = 0.

Test t = pi/2: sin(pi/2) = 1, 1 + cos(pi/2) = 1 + 0 = 1.
This one checks out.

Test t = pi: sin(pi) = 0, 1 + cos(pi) = 1 + (-1) = 0.
This one checks out.

Test t = 3pi/2; sin(3pi/2) = -1, 1 + cos(3pi/2) = 1 + 0 = 1.
Reject t = 3pi/2.

Therefore

t = {pi/2, pi}, or, converted to degrees
t = {90, 180}

Answer 4

cos x/sec x + sin x/csc x = a million secx= a million/cosx cscx=a million/sinx so that you get cosx/(a million/cosx) + sin x/(a million/sin x) = a million cos^2x + sin^2x = a million cos x/ (sec x sin x) = csc x - sin x cosx/ (sinx/cosx) = a million/sinx - sinx cos^2x/sinx = a million/sinx - sinx multiply by sinx to get cos^2x = a million - sin^2x sin^2x + cos ^2x = a million