Grade 10Trigonometry
sin4/a + cos4/b = 1/a+bprove that sin8/a + cos8/b = 1/(a+b)3
sin4/a + cos4/b = 1/a+b
prove that
sin8/a + cos8/b = 1/(a+b)3
1 Answer
Arun
Sin⁴x/a + cos⁴x/b = 1/(a + b)
we know, sin²x + cos²x = 1 ⇒cos²x = 1 - sin²x
cos⁴x = (1 - sin²x)² = 1 + sin⁴x - 2sin²x , use it above
sin⁴x/a + (1 + sin⁴x - 2sin²x)/b = 1/(a + b)
sin⁴x/a + 1/b + sin⁴x/b - 2sin²x/b = 1/(a + b)
sin⁴x(1/a + 1/b) -2sin²x/b = 1/(a + b) - 1/b
sin⁴x(a + b)/ab - 2asin²x/ab = (b - a - b )/(a + b)b
(a + b)²(sin²x)² - 2a(a + b) sin²x = -a²
{(a + b)sin²x}² -2.a.(a + b)sin²x + a² = 0 [ this is like (a - b)² = a² - 2ab + b² ]
{(a + b)sin²x - a}² = 0
sin²x = a/(a + b)
⇒1 - sin²x = cos²x = 1 - a/(a + b) = b/(a + b)
hence, sin²x = a/(a + b) and cos²x = b/(a + b)
so,
sin⁸x = a⁴/(a + b)⁴ and cos⁸x = b⁴/(a + b)⁴
now, sin⁸x/a³ = a/(a + b)⁴
cos⁸x/b³ = b/(a + b)⁴
So, sin⁸x/a³ + cos⁸x/b³ = a/(a + b)⁴ + b/(a + b)⁴ = (a + b)/(a + b)⁴ = 1/(a + b)³
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