Prove the following identities:
1. (tan theta - sin theta)^2 + (1-cos theta)^2 = (1-sec theta) ^2
2. (1-2cos^2 theta) / sin theta cos theta = tan theta - cot theta
3. (sin theta + cos theta ) ^2 + (sin theta - cos theta ) ^2 = 2
Thank you so much! :)
To prove the given identities, we will work on both sides of each equation and simplify them until they are equal.
1. (tan theta - sin theta)^2 + (1-cos theta)^2 = (1-sec theta) ^2
Expanding the left side:
(tan^2 theta - 2tan theta sin theta + sin^2 theta) + (1 - 2cos theta + cos^2 theta) = 1 - 2sec theta + sec^2 theta
Applying trigonometric identities:
(tan^2 theta + sin^2 theta - 2tan theta sin theta) + (1 - 2cos theta + cos^2 theta) = 1 - 2sec theta + sec^2 theta
(1 - 2tan theta sin theta) + (1 - 2cos theta + cos^2 theta) = 1 - 2sec theta + sec^2 theta
2 - 2tan theta sin theta - 2cos theta + cos^2 theta = 1 - 2sec theta + sec^2 theta
Rearranging terms:
2 - 2cos theta - 2tan theta sin theta + cos^2 theta = 1 - 2sec theta + sec^2 theta
(cos^2 theta - 2cos theta + 1) - (2tan theta sin theta + 2sec theta - sec^2 theta) = 0
(cos theta - 1)^2 - 2(sin theta + sec theta) (tan theta - sec theta) = 0
Since (cos theta - 1)^2 is always non-negative, for the equation to be true, the second term (2(sin theta + sec theta) (tan theta - sec theta)) must be equal to zero.
For the second term to be zero:
Either (sin theta + sec theta) = 0 or (tan theta - sec theta) = 0
Case 1: (sin theta + sec theta) = 0
Dividing the equation by sin theta:
1 + cos theta = 0
cos theta = -1
This is true for theta = pi
Case 2: (tan theta - sec theta) = 0
Dividing the equation by tan theta:
1 - sec^2 theta = 0
sec^2 theta = 1
sec theta = 1
This is true for theta = pi/4 or theta = 7pi/4
Therefore, the given identity is true for theta = pi, pi/4, 7pi/4.
2. (1-2cos^2 theta) / sin theta cos theta = tan theta - cot theta
Starting with the left side of the equation:
(1 - 2cos^2 theta) / sin theta cos theta
Applying trigonometric identities:
((1 - cos^2 theta) - cos^2 theta) / sin theta cos theta
(sin^2 theta - cos^2 theta) / sin theta cos theta
(sin^2 theta / sin theta cos theta) - (cos^2 theta / sin theta cos theta)
(tan theta - cos^2 theta / sin theta cos theta)
Now we have tan theta - (cos^2 theta / sin theta cos theta). To simplify further, we use the identity cot theta = cos theta / sin theta to replace (cos^2 theta / sin theta cos theta) with cot theta:
tan theta - cot theta
Therefore, the left side of the equation simplifies to the right side. The identity is proven.
3. (sin theta + cos theta ) ^2 + (sin theta - cos theta ) ^2 = 2
Expanding both squares:
sin^2 theta + 2sin theta cos theta + cos^2 theta + sin^2 theta - 2sin theta cos theta + cos^2 theta = 2
Combining like terms:
2sin^2 theta + 2cos^2 theta = 2
Using the identity sin^2 theta + cos^2 theta = 1:
2(1) = 2
The left side equals the right side, therefore the identity is proven.
Hope this helps! Let me know if you have any other questions.
sinθ = tanθ cosθ, so
tanθ - sinθ = tanθ (1-cosθ)
squared that is tan^2θ (1-cosθ)^2
now we can add up the left side:
tan^2θ(1-cosθ)^2 + (1-cosθ)^2
= (tan^2θ + 1)(1-cosθ)^2
= sec^2θ (1-cosθ)^2
on the right side, we have
1-secθ = (cosθ-1)/cosθ
squared, that is sec^2θ (cosθ-1)^2
and they are the same
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1-2cos^2θ = -cos2θ
2sinθ cosθ = sin2θ
so, you have -2cot2θ
-2cot2θ = -2(1-tan^2θ)/2tanθ
= -(1-tan^2θ)/tanθ
= -(cotθ - tanθ)
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(sinθ+cosθ)^2 = sin^2θ + 2sinθcosθ + cos^2θ
expand the other one, then add and it drops right out.