To prove the first trigonometry identity, we start with the left side of the equation:
(1 β cos^2ΞΈ) sinΞΈ cos^2ΞΈ
Using the Pythagorean identity sin^2ΞΈ + cos^2ΞΈ = 1, we can rewrite cos^2ΞΈ as:
1 β sin^2ΞΈ
Substituting this back into the previous equation, we get:
(1 β cos^2ΞΈ) sinΞΈ (1 β sin^2ΞΈ)
Expanding this multiplication gives:
sinΞΈ β sin^3ΞΈ
Now, using the identity sin^3ΞΈ = sinΞΈ(1 β cos^2ΞΈ), which can be derived from sin^2ΞΈ = 1 β cos^2ΞΈ, we can rewrite the previous equation as:
sinΞΈ β sinΞΈ(1 β cos^2ΞΈ)
Simplifying further gives:
sinΞΈ β sinΞΈ + sinΞΈcos^2ΞΈ
Combining like terms, we obtain:
sinΞΈcos^2ΞΈ
Since sinΞΈcos^2ΞΈ is equal to (1/2)sinΞΈ(2cos^2ΞΈ), which is equal to (1/2)(2sinΞΈcosΞΈcosΞΈ) by using the double angle identity, we can rewrite sinΞΈcos^2ΞΈ as:
(1/2)sin2ΞΈcosΞΈ
Using the identity sin2ΞΈ = 2sinΞΈcosΞΈ, we get:
(1/2)(2sinΞΈcosΞΈ)cosΞΈ
Simplifying further:
(1/2)(2sinΞΈcos^2ΞΈ)
Now, using the identity cos^2ΞΈ = 1 β sin^2ΞΈ, we can rewrite cos^2ΞΈ as:
(1/2)(2sinΞΈ(1 β sin^2ΞΈ))
Expanding the multiplication gives:
sinΞΈ β sin^3ΞΈ
This is equal to the right side of the equation, thus proving the identity:
(1 β cos^2ΞΈ)sinΞΈcos^2ΞΈ = 1
Prove the following trigonometry identities
1. (1 β πππ 2π)ππ π2π = 1
2.π πππβ(1 β π ππ2π) = 1 [5]
3. πππ π
(1βπ‘πππ)
+
π πππ
1βπ0π‘π = π πππ + πππ π
1 answer