i have no clue how to do identities. I have the answers so just giving that does not help at all, can anyone actually explain what to do with any of these problems???
find the exact value of: cos^2(75) -- sin^2(75)
cos(73) cos(17) -- sin(73) sin(17)
find the exact value of: cos^2(75) -- sin^2(75)
cos(73) cos(17) -- sin(73) sin(17)
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You're just going to have to learn some trig identities, and use them enough that you can recognize which one(s) to use for a given problem. I realize you probably won't be able to do that in one night.
For the first one, you need the identity cos(2x) = cos²(x) - sin²(x), so set x = 75°.
cos(150°) = cos²(75°) - sin²(75°)
The terminal side of an angle of measure 150° lies in the second quadrant, and hence its cosine is negative. Its reference angle is 30². Recall that an angle and its reference angle have the same trig values, except possibly for sign. cos(30°) = (√3)/2, so cos(150°) = -√3/2. Therefore, cos²(75°) - sin²(75°) = -(√3)/2 .
For the second one, you need the identity cos(x + y) = cos(x) cos(y) - sin(x) sin(y).
For your problem, x = 73° and y = 17°, so
cos(73° + 17°) = cos(73°) cos(17°) - sin(73°) sin(17°)
cos(90°) = cos(73°) cos(17°) - sin(73°) sin(17°)
0 = cos(73°) cos(17°) - sin(73°) sin(17°)
Other identities you should know:
sin(2x) = 2 sin(x) cos(x)
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
sin(x - y) = sin(x) cos(y) - cos(x) sin(y)
cos(x - y) = cos(x) cos(y) + sin(x) sin(y)
Note that the sin(2x) formula is a special case of the second identity, with y = x. Similarly, the identity used in the first problem is a special case of the identity used in the second problem, again with y = x.
Good luck!
For the first one, you need the identity cos(2x) = cos²(x) - sin²(x), so set x = 75°.
cos(150°) = cos²(75°) - sin²(75°)
The terminal side of an angle of measure 150° lies in the second quadrant, and hence its cosine is negative. Its reference angle is 30². Recall that an angle and its reference angle have the same trig values, except possibly for sign. cos(30°) = (√3)/2, so cos(150°) = -√3/2. Therefore, cos²(75°) - sin²(75°) = -(√3)/2 .
For the second one, you need the identity cos(x + y) = cos(x) cos(y) - sin(x) sin(y).
For your problem, x = 73° and y = 17°, so
cos(73° + 17°) = cos(73°) cos(17°) - sin(73°) sin(17°)
cos(90°) = cos(73°) cos(17°) - sin(73°) sin(17°)
0 = cos(73°) cos(17°) - sin(73°) sin(17°)
Other identities you should know:
sin(2x) = 2 sin(x) cos(x)
sin(x + y) = sin(x) cos(y) + cos(x) sin(y)
sin(x - y) = sin(x) cos(y) - cos(x) sin(y)
cos(x - y) = cos(x) cos(y) + sin(x) sin(y)
Note that the sin(2x) formula is a special case of the second identity, with y = x. Similarly, the identity used in the first problem is a special case of the identity used in the second problem, again with y = x.
Good luck!