Trigonometry, different products of sine and cosine

AI Thread Summary
In a right-angled triangle with angles a, b, and 90 degrees, where a < b, the expressions sin a sin b, sin a cos b, cos a sin b, and cos a cos b yield three distinct values. The key relationship is that sin(a) equals cos(b) and sin(b) equals cos(a), ensuring that the two angles are not equal. This distinction is crucial because if a were equal to b, all expressions would yield the same value. To visualize this, drawing a right triangle and substituting specific side lengths can clarify the relationships between the trigonometric functions. Understanding these identities and their implications is essential for solving the problem effectively.
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Homework Statement



There is a right angled triangle, with the following angles, a, b, and 90deg.

If a < b, how many different values are there among the following expressions?

sin a sin b, sin a cos b, cos a sin b, cos a cos b

Homework Equations





The Attempt at a Solution



I don't really know any trigonometric identies and I am guessing that's where the solution lies :S
 
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Here is a trigonometric identity: sin(x) = cos(90 - x)
Try to use it.
 
:S

sorry i still don't know how i would use it
 
What is the sum of the 3 angles in a triangle?
And if you know that one of them is a right angle, then what is the sum of the other 2?
 
Trail_Builder said:

Homework Statement



There is a right angled triangle, with the following angles, a, b, and 90deg.

If a < b, how many different values are there among the following expressions?

sin a sin b, sin a cos b, cos a sin b, cos a cos b

Homework Equations





The Attempt at a Solution



I don't really know any trigonometric identies and I am guessing that's where the solution lies :S

antonantal said:
Here is a trigonometric identity: sin(x) = cos(90 - x)
Try to use it.

Trail_Builder said:
:S

sorry i still don't know how i would use it
The point is that sin(a)= cos(b) and sin(b)= cos(a).
 
there are 3 different values there. the oiint in telling you that b>a is to make sure you know that b does not equal to a. because, is b=a, the values for all the expressions will be the same. but in this case, since the 2 angles are different, sina=cosb. also, sinb=cosa. if you don't get this, try drawing out a right-angled triangle and label it's angles. use 3,4,5 as the length of its sides. 5 is the hypothenus. then sub the values into the expressions. you will find that sina sinb=cosa cosb.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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