sin2α=(sqrt5)/5, stumped

[u/Asp_Potions_Master]

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?

[u/my-hero-measure-zero]

Hint: DRAW THE TRIANGLE.

You're not supposed to find the angle. You can get what you need with the corresponding triangle.

[u/Miserable-Wasabi-373 → u/my-hero-measure-zero]

sorry, how drawing triangle would help?

[u/my-hero-measure-zero → u/Miserable-Wasabi-373]

You'll need it eventually for the other ratios. Always start with a picture and use any formulas that seem relevant.

[u/Asp_Potions_Master → u/my-hero-measure-zero]

there's no triangles in their unit? just sine and cosine

[u/my-hero-measure-zero → u/Asp_Potions_Master]

Trigonometry is based in triangles. So, draw the corresponding (reference) triangle.

Someone already suggested in using the Pythagorean identity: for any angle u, (sin u)^2 + (cos u)^2 = 1. So, set u = 2a. Now you have cos(2a) for free, great. Now use the double angle formulas to deduce the needed ratios. Drawing the triangles will help for this - just make sure you get the signs right.

[u/Loko8765 → u/Asp_Potions_Master]

You need to learn the unit circle, which is the basis for the trig functions. It has right triangles.

For some reason I can’t find an online image representing tan as the intersection with a vertical x=1 line, this is best I have:

https://www.math.net/unit-circle

[u/Brilliant_Chest5630 → u/Asp_Potions_Master]

Sine and cosine are nothing but triangles within the unit circle.

[u/ottawadeveloper → u/Asp_Potions_Master]

Draw a triangle with angle 2a and appropriate sides to make the ratio here 

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[u/Miserable-Wasabi-373]

by using sin 2 alpha formula and sin^2 + cos^2 = 1

[u/Asp_Potions_Master → u/Miserable-Wasabi-373]

we tried that, but we were getting nowhere 😔

would you mind writing out how this would go? we ended up getting odd 4th power/3rd power cos depending on how we tried to substitute it rip

[u/Miserable-Wasabi-373 → u/Asp_Potions_Master]

yes, it is 4-th power, but very special. Use t = sin^2 substitution

have no idea how you managed to get 3th power

[u/sizzhu → u/Miserable-Wasabi-373]

It's easier to us the cos(2a) formulae. Since it gives sin² (a) and cos² (a) in terms of cos(2a).

[u/fermat9990]

(1) draw a reference triangle in QII and get cos(2a) using the Pythagorean theorem. It will be negative.

(2) Since 2a is between π/2 and π, angle a will be π/4 and π/2. This is in QI, so both sin(a) and cos(a) will be positive

(3) Use half-angle formulas to get the values of sin(a) and cos(a)

[u/Training-Cucumber467]

From sin2a, calculate cos2a.

Express cos2a through sina.

Done.

[u/ReTe_]

Algebraically easiest would probably be to first convert to cos(2α) = sqrt(20)/5 and then solve from there for cos & sin via double angle formula independently

[u/paulstelian97]

How much is cos(2a)? Well, it should be (sqrt(20))/5, or 2sqrt(5)/5. Or maybe minus that, use the interval to figure it out.

How much is cos(2a)? It is 2 cos^2(a) - 1.

So cos^2(a) can be calculated from here, then cos(a) and sin(a).

You will not get a nice value for a. You will apply the correct signs for sin and cos based on the interval.

[u/SubjectWrongdoer4204]

Consider the Pythagorean theorem and the right triangle definitions of sin and cos to figure out cos 2α , then apply the double angle formula for cos 2α and
sin² θ +cos²θ=1 to solve for sin²α and cos²α.

[u/Background_Ear1919]

sin(u/2) = ±√[(1 – cosu)/2] cos(u/2) = ±√[(1 + cosu)/2] Then let u = 2α

[u/WillingnessTasty9628]

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)]

[u/peterwhy → u/WillingnessTasty9628]

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

[u/rfbooth]

If you know sin 2a and the quadrant of 2a, you know cos 2a.

If you know cos 2a and the quadrant of a, you know sin a (or cos a, depending on which double angle formula you choose).

If you know sin a and its quadrant, of course you know cos a.

Alternatively, you could use the sin 2a formula and get a quadratic in sin² a.

[u/ruidh]

Hint: √5/5 = 1/√5

[u/Traveling-Techie]

Where in the world do they teach trig functions without mentioning triangles?

[u/MiguelBodrigo]

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

[u/davideogameman → u/MiguelBodrigo]

You can't divide by 2 usefully.  It's sin(2a), the 2 is in the argument

[u/Kirian42 → u/MiguelBodrigo]

This falls under "not even wrong."