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Zorluk: OrtaFundamental Trigonometric Identities

If sin⁡θ−cos⁡θ=0.6\sin \theta - \cos \theta = 0.6, what is the value of sin⁡θcos⁡θ\sin \theta \cos \theta?

Cevap: 0.32

Cevap

The value of sin⁡θcos⁡θ\sin \theta \cos \theta is 0.32.
Squaring both sides of sin⁡θ−cos⁡cos⁡θ=0.6\sin \theta - \cos \cos \theta = 0.6 yields sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36. Substituting the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 gives 1−2sin⁡θcos⁡θ=0.361 - 2\sin \theta \cos \theta = 0.36, which rearranges to 2sin⁡θcos⁡θ=0.642\sin \theta \cos \theta = 0.64. Dividing by 2 yields sin⁡θcos⁡θ=0.32\sin \theta \cos \theta = 0.32.

Adım Adım Çözüm

1
Square both sides of the given equation
(sin⁡θ−cos⁡θ)2=0.36(\sin \theta - \cos \theta)^2 = 0.36
Squaring allows us to introduce the product sin⁡θcos⁡θ\sin \theta \cos \theta alongside sin⁡2θ\sin^2 \theta and cos⁡2θ\cos^2 \theta.
2
Expand the binomial on the left side
sin⁡2θ−2sin⁡θcos⁡θ+cos⁡2θ=0.36\sin^2 \theta - 2\sin \theta \cos \theta + \cos^2 \theta = 0.36
Use the algebraic expansion identity (a−b)2=a2−2ab+b2(a - b)^2 = a^2 - 2ab + b^2.
3
Substitute the fundamental Pythagorean trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
1−2sin⁡θcos⁡θ=0.361 - 2\sin \theta \cos \theta = 0.36
sin⁡2θ+cos⁡2θ\sin^2 \theta + \cos^2 \theta always equals 1 for any angle θ\theta.
4
Isolate the term containing sin⁡θcos⁡θ\sin \theta \cos \theta
2sin⁡θcos⁡θ=0.642\sin \theta \cos \theta = 0.64
Subtract 0.36 from 1 to find the value of 2sin⁡θcos⁡θ2\sin \theta \cos \theta.
5
Divide by 2 to solve for sin⁡θcos⁡θ\sin \theta \cos \theta
sin⁡θcos⁡θ=0.32\sin \theta \cos \theta = 0.32
Simplifies 0.64/20.64 / 2 to obtain the final required numerical value.

Anahtar Kavram

Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
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Fundamental Trigonometric Identities

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