Sin cos problem finding exact value

[aisha]

sin cos problem finding exact value...

I have to find the exact value of : [tex] \frac {\sin^2 45 degrees + \cos ^2 45 degrees} {\sin 60 degrees \cos 30 degrees}[/tex]

The squares are throwing me off , without them I got
[tex] \frac {\frac {1} {sqrt2} + \frac {1} {sqrt2}} {3/2} [/tex]

what do I do with the sqares on sin and cos in the numerator?

 

[thursdaytbs]

Try this:

[tex] sin^2 45 degrees [tex], Sin(45) = radical 2 / 2, so sin^2(45) = 2/4

 

[dextercioby]

If u use these formulas
[tex] \sin^{2}x+\cos^{2}x=1[/tex]
,for all "x"
and
[tex] \sin(90-x)=\cos x [/tex]
,u'd have the problem solved.

Besides,your calculations are wrong,since at the numerator they didn't include squaring thevalues for sine and cosine,which were supposed to be done.

And the denominator is 3/4.

Daniel.

 

[aisha]

ok without the square roots on the sine and cosine I get 2/3 but the solution says the answer is 4/3?

 

[dextercioby]

If the denominator is 3/4 and the numerator is 1,what is the whole fraction equal to??

Daniel.

 

[aisha]

[tex] \sin 60 = \frac {sqrt3} {2} [/tex]

and doesn't [tex] \cos 30 = \frac {sqrt3} {2} [/tex] also?

these multiplied gives 3/2 ??

How do u get the denominator to equal 3/4?

 

[dextercioby]

The multiplication is wrong.U multiplied only the numerators.U need to multiply the denominators as well.I'm sure you'll get
[tex] 2\cdot 2 =4[/tex]

Daniel.

 

[thursdaytbs]

(sqrt3/2) * (sqrt3/2) = 3/4

you forgot to multiply the two TWO's.

I've done that before, =\.

 

[aisha]

(sqrt3/2) * (sqrt3/2) = 3/4

you forgot to multiply the two TWO's.

I've done that before, =\.

LOL omg silly me Thanks Holy can't believe I missed that thanks