Question

cos4A -sin4A + 1 = 2cos2A
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Answer 1

Step-by-step explanation:

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Solved Examples on Trigonometry

Example1: sin θ cos 7θ cos 4θ + sin 7θ cos θ cos 4θ + sin 4θ cos 7θ cos θ = sin θ sin 4θ sin 7 θ.

Solution: This question is already in sine and cosine ratios. This preferable form, but this does not seem to be of much help.

Moreover this is not in terms of one single Trigonometric Ratio i.e. either sine or cosine. Hence it is not easy to proceed in the present form. What shall we do? Divide whole thing by cos θ cos 4θ cos7θ so that we get the entire expression in terms of tangent ratios,

tan θ + tan 7θ + tan 4θ = tan θ tan 4θ tan 7θ

tan θ + tan 4θ = – tan 7θ (1 – tan θ tan 4θ).

Case I:

If 1 – tan θ tan 4θ ≠ 0, you can peacefully divide by this expression.

(tan θ + tan 4θ)/(1 – tan θ tan 4θ) = –tan 7θ

or, tan 5θ = – tan 7θ

⇒ 5θ = nπ – 7θ

⇒ θ = (nπ)/12, where n = 0, ±1, ±2 ………

Check, whether this is a solution and it satisfies the assumption 1 – tan θ tan 4θ ≠ 0.

1 – tan 4 (nπ/12). tan (nπ/12)

1 – tan (nπ/3) tan (nπ/12) ≠ 0

Case II:

1 – tan θ tan 4θ = 0, then given equation becomes

cos θ cos 4θ – sin θ sin 4θ = 0

or, (cos 3θ – cos 5θ) – (cos 5θ – cos 3θ) = 0 or,

2 (cos 3θ – cos 5θ) = 0

or, cos 3θ = cos 5θ

⇒ 5θ = 2nπ ± 3θ ……… (1) t

Taking positive sign of (1) or, 2θ = 2nπ

or, θ = nπ, where n = 0, ±1, ±2 ………

Taking negative sign of (1) or, 8θ = 2nπ

or, θ = nπ/4 where n = 0, ±1, ±2 ………

Check

Put θ = nπ in given equation 0 = 0

Put θ = nπ/4 in given equation 1/√2.1/√2 (-1) + (1/√2)(1/√2)(-1) + 0 = 0

So θ = nπ and nπ/4. Also satisfy given equation θ = nπ, nπ/4, nπ/12, where n = 0, ±1, ±2 ………