Prove that \(\frac{(sin 7x+sin 5x)+(sin 9x+sin3x)}{cos 7x+cos5x)+(cos9x+cos3x)}=tan\,6x.\)
Solution and Explanation
It is known that
\(sinA+sin B=2sin(\frac{A+B}{2}).cos(\frac{A-B}{2}),cosA+cosB=2cos(\frac{A+B}{2}).cos(\frac{A-B}{2})\)
L.H.S. = \(\frac{(sin 7x+sin 5x)+(sin 9x+sin3x)}{(cos 7x+cos5x)+(cos9x+cos3x)}\)
\(\frac{[2sin(\frac{7x+5x}{2}).cos(\frac{7x-5x}{2})]+[2sin(\frac{9x+3x}{2}.cos(\frac{9x-3x}{2})]}{[2cos(\frac{7x+5x}{2}.cos(\frac{7x-5x}{2})]+[2cos(\frac{9x+3x}{2}).cos(\frac{9x-3x}{2})]}\)
\(=\frac{[2sin6x.cosx]+[2sin6x.cos3x]}{[2cos6x.cosx]+[2cos6x.cos3x]}\)
\(=\frac{2sin6x[cosx+cos3x]}{2cos6x[cosx+cos3x]}\)
\(= tan\, 6x\)
\(= R.H.S.\)
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Concepts Used:
Trigonometric Functions
The relationship between the sides and angles of a right-angle triangle is described by trigonometry functions, sometimes known as circular functions. These trigonometric functions derive the relationship between the angles and sides of a triangle. In trigonometry, there are three primary functions of sine (sin), cosine (cos), tangent (tan). The other three main functions can be derived from the primary functions as cotangent (cot), secant (sec), and cosecant (cosec).
Six Basic Trigonometric Functions:
- Sine Function: The ratio between the length of the opposite side of the triangle to the length of the hypotenuse of the triangle.
sin x = a/h
- Cosine Function: The ratio between the length of the adjacent side of the triangle to the length of the hypotenuse of the triangle.
cos x = b/h
- Tangent Function: The ratio between the length of the opposite side of the triangle to the adjacent side length.
tan x = a/b
Tan x can also be represented as sin x/cos x
- Secant Function: The reciprocal of the cosine function.
sec x = 1/cosx = h/b
- Cosecant Function: The reciprocal of the sine function.
cosec x = 1/sinx = h/a
- Cotangent Function: The reciprocal of the tangent function.
cot x = 1/tan x = b/a
Formulas of Trigonometric Functions:
