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.
The point is that sin(a)= cos(b) and sin(b)= cos(a).Trail_Builder said::S
sorry i still don't know how i would use it
Trigonometry is a branch of mathematics that deals with the study of triangles and the relationships between their sides and angles.
The different products of sine and cosine include the sine of an angle multiplied by the cosine of the same angle (sin x * cos x), the sine of an angle multiplied by the sine of the same angle (sin x * sin x), and the cosine of an angle multiplied by the cosine of the same angle (cos x * cos x).
These products are useful in trigonometry because they can help solve problems involving trigonometric identities and equations. They also play a key role in the applications of trigonometry in fields such as physics, engineering, and astronomy.
One example of a problem involving products of sine and cosine is finding the value of sin x * cos x given that sin x = 3/5 and cos x = 4/5. To solve this problem, we can use the identity sin x * cos x = (1/2) * sin 2x. Substituting the given values, we get sin 2x = (3/5)(4/5) = 12/25. Then, we can use the double angle formula sin 2x = 2sin x * cos x to find the value of sin x * cos x = (12/25)/2 = 6/25.
Yes, there are several other important trigonometric identities involving products of sine and cosine, such as the double angle formula for cosine (cos 2x = cos^2 x - sin^2 x) and the half angle formulas for sine and cosine (sin (x/2) = ± √[(1 - cos x)/2] and cos (x/2) = ± √[(1 + cos x)/2]). These identities are frequently used in trigonometric calculations and problem-solving.