Sin cos problem finding exact value

  • Thread starter Thread starter aisha
  • Start date Start date
  • Tags Tags
    Cos Sin Value
Click For Summary

Homework Help Overview

The discussion revolves around finding the exact value of a trigonometric expression involving sine and cosine functions, specifically \(\frac {\sin^2 45^\circ + \cos^2 45^\circ} {\sin 60^\circ \cos 30^\circ}\). Participants are exploring the implications of squaring the sine and cosine values in the numerator and how that affects the overall calculation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the correct application of trigonometric identities and the implications of squaring the sine and cosine values. There are attempts to clarify the calculations for both the numerator and denominator, with some participants questioning the correctness of previous calculations.

Discussion Status

The discussion is active, with various participants providing insights and corrections regarding the calculations. Some guidance on using trigonometric identities has been suggested, but there is no explicit consensus on the final value of the expression.

Contextual Notes

Participants are navigating through potential miscalculations and assumptions regarding the values of sine and cosine, particularly in relation to their squared forms. There is an ongoing exploration of how these values contribute to the overall fraction.

aisha
Messages
584
Reaction score
0
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?
 
Last edited:
Physics news on Phys.org
Try this:

[tex]sin^2 45 degrees [tex], Sin(45) = radical 2 / 2, so sin^2(45) = 2/4[/tex][/tex]
 
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.
 
ok without the square roots on the sine and cosine I get 2/3 but the solution says the answer is 4/3?
 
If the denominator is 3/4 and the numerator is 1,what is the whole fraction equal to??

Daniel.
 
[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?
 
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.
 
(sqrt3/2) * (sqrt3/2) = 3/4

you forgot to multiply the two TWO's.

I've done that before, =\.
 
thursdaytbs said:
(sqrt3/2) * (sqrt3/2) = 3/4

you forgot to multiply the two TWO's.

I've done that before, =\.

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

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 8 ·
Replies
8
Views
1K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
Replies
46
Views
8K
Replies
4
Views
1K