ACT Math 1.3 Trigonometry
ACT Math: Trigonometry Drill 1, Problem 3. Can you find the cosine?
|Foreign Language||Arabic Subtitled|
|Geometry||Right Triangles and Trigonometry|
|Product Type||ACT Math|
Values and properties of trigonometric functions
A times cosine B plus sine A
time sine B. Now have to decide what the values A and B are.
What are two friendly angles?
As in nice trig values that have a different of 105
150 - 45 is 105, though A equals 150 and B equals
45. Now I can plug it into the difference formula for cosine.
We can figure out what cosine of 150 degrees is based on the cosine of
30 degrees. The 150 degrees is just a mulitple of 30 degrees
but in the second quadrant or cosine is negative...the cosine of 150
equals negative cosine 30 degrees. Cosine of 30 degrees equal square root of
3 over 2
so cosine to 150 degrees is negative root 3
over 2. Cosine of 45 degrees is the square root of two over two.
We can find sine of 150 degrees based on sine of
30 degrees because once again a 150 is a multiple of 30 degrees
since sin is still positive in Quadrant 2...
...Sine 150 degrees equals the same value sine of thirty degrees.
Sine 30 is one half and so is sin 150
Sine 45 is the square root of 2 over 2
we have negative square root three over two times the square root
2 over two plus one half time through 2 over 2
Simplified we get negative square root of six over four-plus the square root of
two over 4
which equals root 2 minus root 6 all over four.
Option C is our answer.