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# Double Angle Identities Calculus Help?

My teacher wasn't here today and left us new work. I did most of it expect three

Prove the identity- cos6x = 2cos^2 3x-1 (^2 means raised to the 2nd power)

Solve algebraically for exact solutions- cos2x + sinx =0 & cosx + cos3x = 0

Please show all work thanks you!!

### 1 Answer

- Anonymous10 years agoFavorite Answer
1. LHS . . . . RHS

cos(6x) = 2cos²(3x) - 1

* * *

Rewrite cos(6x) as:

cos(2(3x))

Recall that:

cos(2x) = 2cos²(x) - 1

= 1 - 2sin²(x)

= cos²(x) - sin²(x)

Finally, by double-angle identity (letting x = 3x), we obtain:

2cos²(3x) - 1

==> RHS

2. cos(2x) + sin(x) = 0

By double-angle identity, we obtain:

1 - 2sin²(x) + sin(x) = 0

-2sin²(x) + sin(x) + 1 = 0

Then,

(2sin(x) + 1)(-sin(x) + 1) = 0

2sin(x) + 1 = 0 and -sin(x) + 1 = 0

sin(x) = -½ and sin(x) = 1

x = arcsin(-½) and x = arcsin(1)

x = -π/6, π/2

Since the period is 2π, we obtain:

x = -π/6 + 2n, π/2 + 2πn, where n is the integer

3. cos(x) + cos(3x) = 0

Recall that:

cos(3x) = 4cos³(x) - 3cos(x)

By triple-angle identity, we obtain:

cos(x) + 4cos³(x) - 3cos(x) = 0

4cos³(x) - 2cos(x) = 0

Then,

2cos(x)(2cos²(x) - 1) = 0

2cos(x) = 0 and 2cos²(x) - 1 = 0

cos(x) = 0 and cos(x) = ±√(½)

cos(x) = 0 and cos(x) = √(½) and cos(x) = -√(½)

x = arccos(0) and x = arccos(√(½)) and x = arccos(-√(½))

Therefore,

x = πn - π/2 where n is the integer

I hope this helps!

Source(s): Knowledge