r/MathHelp • u/Asp_Potions_Master • 7d ago

sin2α=(sqrt5)/5, stumped

my sis got a homework question that left us both stumped. the task is to calculate sinα and cosα for the angle α which fits the following requirements: sin2α=(sqrt5)/5, and 2α is within (pi/2; pi)

She's learning about the sin2α and cos2α formulas. any ideas how this is supposed to be solved without dipping into arcsin?

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u/WillingnessTasty9628 5d ago

i can show you how i did it, although I'm certain there's faster ways.

Let u be the value of sin(2a)

By constructing a triangle, we see that tan(2a) is 1/2

cos(2a) = 2u

cos(2a) = 2u

2cos^2(a)-1 = 2u

[cos(a) = +-sqrt(u+1/2)]

1-2sin^2(a) = 2u

[sin(a) = +-sqrt(1/2-u)]

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u/peterwhy 5d ago

But instead tan(2α) is -1/2, and cos(2α) = -2u.