Prove that each equation is an identity.
I tried to do the problems, but I am stuck.
1. cos^4 t-sin^4 t=1-2sin^2 t
2. 1/cos s= csc^2 s - csc s cot s
3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x
4. sin^3 z cos^2 z= sin^3 z - sin^5 z
cos^4 - sin^4 = (cos^2+sin^2)(cos^2-sin^2)
Remember algebra I and your double-angle formulas?
working with the right side, we have
csc^2 - csc cot
1/sin^2 (1 - cos)
1/(1-cos^2) (1-cos)
(1-cos) / (1+cos)(1-cos)
1/(1+cos)
Hmmm. I don't get 1/cos
(cos x/ (sec x -1))- (cos x/ tan^2x)
(cos/(1/cos-1))-cos^3/sin^2
cos(cos-1) - cos^3/sin^2
cos^2-1 - cos^3/sin^2
-1/sin^2 (sin^4 - cos^3)
Hmmm. I don't see how to make that
cos^2/sin^2
excuse me?
cos^2 = 1-sin^2
and it drops right out
To prove that each equation is an identity, we need to simplify both sides of the equation until we reach the same expression on both sides. Let's go through each equation step by step.
1. cos^4 t - sin^4 t = 1 - 2sin^2 t
Start with the left-hand side:
cos^4 t - sin^4 t
This can be factored as a difference of squares:
(cos^2 t - sin^2 t)(cos^2 t + sin^2 t)
Recall the Pythagorean identity: cos^2 t + sin^2 t = 1
Substitute this into the equation:
(cos^2 t - sin^2 t)(1) = 1 - 2sin^2 t
Use the difference of squares formula: a^2 - b^2 = (a + b)(a - b)
[(cos t + sin t)(cos t - sin t)] = 1 - 2sin^2 t
Cos t - sin t is the same as cos (-t) + sin (-t) by an angle addition formula.
(cos t + sin t)(cos (-t) + sin (-t)) = 1 - 2sin^2 t
Apply the commutative property of multiplication to rearrange the terms:
(cos t + sin t)(sin t + cos t) = 1 - 2sin^2 t
The left-hand side of the equation can be simplified using the distributive property:
sin^2 t + cos^2 t + sin t cos t + sin t cos t = 1 - 2sin^2 t
Using the Pythagorean identity again, cos^2 t + sin^2 t = 1:
1 + 2sin t cos t = 1 - 2sin^2 t
Simplify further by subtracting 1 from both sides:
2sin t cos t = -2sin^2 t
Divide both sides by -2:
sin t cos t = -sin^2 t
Now we can use the Pythagorean identity sin^2 t = 1 - cos^2 t:
sin t cos t = -(1 - cos^2 t)
Multiply through by -1 to simplify:
- sin t cos t = 1 - cos^2 t
Finally, substitute cos^2 t = 1 - sin^2 t:
- sin t cos t = 1 - (1 - sin^2 t)
Simplify the equation further:
- sin t cos t = sin^2 t
For the right-hand side, rewrite sin^2 t as sin t*sin t:
- sin t cos t = sin t*sin t
Thus, we have proved that cos^4 t - sin^4 t = 1 - 2sin^2 t is an identity.
You can follow a similar approach to prove the other equations are identities.