Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

How do you write asinx+bcozx as a single trigonometric ratio?please show all steps.?

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• 1 month ago

Consider the right triangle in the diagram below.

In the right triangle, θ is an angle with opposite side b and adjacent side a.

By Pythagorean theorem, the hypotenuse = √(a² + b²)

Hence, cos(θ) = a/√(a² + b²) and sinθ = b/√(a² + b²)

Also tan(θ) = b/a, and thus θ = tan⁻¹(b/a)

Besides, consider the following trigonometric identity:

sin(A + B) = cos(B) sin(A) + sin(B) cos(A)

a sin(x) + b cos(x)

= √(a² + b²) * {[a/√(a² + b²)] sin(x) + [b/√(a² + b²)] cos(x)}

= √(a² + b²) * {cos(θ) sin(x) + sin(θ) cos(x)}

= √(a² + b²) sin[x + θ]

= √(a² + b²) sin[x + tan⁻¹(b/a)]

• 1 month ago

Think of a and b as the legs of another right triangle.  I'm sure you meant a * sin(x) + b * cos(x) as well.

sqrt(a^2 + b^2) * (a * sin(x) / sqrt(a^2 + b^2)  +  b * cos(x) / sqrt(a^2 + b^2))

a / sqrt(a^2 + b^2) = cos(t)

b / sqrt(a^2 + b^2) = sin(t)

sqrt(a^2 + b^2) * (sin(x) * cos(t) + sin(t) * cos(x))

sqrt(a^2 + b^2) * sin(x + t)

• Nate1 month agoReport

so is that the same as say, Rsin(x+t)?