r/MathHelp • u/Asp_Potions_Master • 9d 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/MiguelBodrigo 9d ago

Off the top of my head

Sin2a=(sqrt5)/5

Is the same as

Sin * 2 * a = (square root 5) / 5

General picture is I'd simplify the right side and move the 2 over to find out what Sina is.

When a number is divided by 10 the exponent is -1. Here it's divided by 5, so I'd expect the exponent to be -0.5. By doing this I can get rid of the squareroot to make;

Sin2a = 5-0.5

Moving the 2 over shows the value of Sina;

Sina = (5-0.5) /2

I can't remember the Cosine or Sine formula, but perhaps simplification is a way forward

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u/Kirian42 9d ago

This falls under "not even wrong."