Question

Sin⁴a+Cos⁴a=1-0.5Sin²×2a

Answer 1

I hope this helps you

Answer 2

Let's prove the given trigonometric identity step by step:

Starting with the left-hand side (LHS): LHS = sin^4(a) + cos^4(a)

Now, we know the following trigonometric identities:

1. sin^2(a) + cos^2(a) = 1
2. sin^2(2a) + cos^2(2a) = 1

We can use these identities to rewrite the LHS in terms of sin^2(a) and cos^2(a): LHS = (sin^2(a))^2 + (cos^2(a))^2

Next, we can apply identity #1 to the above expression: LHS = (1 - cos^2(a))^2 + (cos^2(a))^2

Expanding the squared terms: LHS = (1 - 2cos^2(a) + cos^4(a)) + (cos^2(a))^2

Simplifying: LHS = 1 - 2cos^2(a) + cos^4(a) + cos^4(a)

Now, we can combine like terms: LHS = 1 + 2cos^4(a) - 2cos^2(a)

Now, let's work on the right-hand side (RHS): RHS = 1 - 0.5sin^2(2a)

Using identity #2, we can express sin^2(2a) in terms of sin^2(a) and cos^2(a): sin^2(2a) = 2sin^2(a)cos^2(a)

Now, we can substitute this into the RHS: RHS = 1 - 0.5(2sin^2(a)cos^2(a))

RHS = 1 - sin^2(a)cos^2(a)

Now, we can use the identity sin^2(a)cos^2(a) = (sin^2(a))(1 - sin^2(a)): RHS = 1 - (sin^2(a))(1 - sin^2(a))

Using the identity sin^2(a) + cos^2(a) = 1, we can simplify further: RHS = 1 - (sin^2(a))(cos^2(a))

Now, we can use the double-angle identity for sine: sin(2a) = 2sin(a)cos(a) sin^2(a) = (1/2)sin(2a) (dividing both sides by 2)

Substituting this into the RHS: RHS = 1 - ((1/2)sin(2a))(cos^2(a))

Using the double-angle identity for cosine: cos(2a) = cos^2(a) - sin^2(a) cos^2(a) = cos(2a) + sin^2(a)

Substituting this into the RHS: RHS = 1 - ((1/2)sin(2a))(cos(2a) + sin^2(a))

Expanding the multiplication: RHS = 1 - (1/2)sin(2a)cos(2a) - (1/2)sin^3(a)

Now, we can use the double-angle identity for sine: sin(2a) = 2sin(a)cos(a) RHS = 1 - (1/2)(2sin(a)cos(a))(cos(2a)) - (1/2)sin^3(a)

Simplifying: RHS = 1 - sin(a)cos(2a) - (1/2)sin^3(a)

Finally, using the double-angle identity for cosine: cos(2a) = 1 - 2sin^2(a) RHS = 1 - sin(a)(1 - 2sin^2(a)) - (1/2)sin^3(a)

Expanding the multiplication: RHS = 1 - sin(a) + 2sin^3(a) - (1/2)sin^3(a)

Combining like terms: RHS = 1 - sin(a) + (3/2)sin^3(a)

Now, comparing the LHS and RHS: LHS = 1 + 2cos^4(a) - 2cos^2(a) RHS = 1 - sin(a) + (3/2)sin^3(a)

It appears that the expressions on the LHS and RHS are not equal to each other. The trigonometric identity you provided, sin^4(a) + cos^4(a) = 1 - 0.5sin^2(2a), is not valid.