I can't find the exact value for this equation. Please help me!
2006-12-11 15:38:54 · 6 answers · asked by keenan93312 1 in Science & Mathematics ➔ Mathematics
sin^-1(-2/7) is an angle in a right triangle whose hypotenuse is 7 and side opposite is -2. The cosine is side adjacent over the hypotenuse, so you need to find the value of that adjacent side. From the Pythagorean theorem, a = √[(7^2 - (-2)^2] = √[49-4=√45. The value of the cosine is then x = (√45)/7. Since the value of the sine is negative, the angle must be in the third quadrant or fouth quadrant. The cosine in those quadrants can be either + or -, so the answer is ±(√45)/7
2006-12-11 15:49:32 · answer #1 · answered by gp4rts 7 · 2⤊ 0⤋
Since b=-2 and c=7, you do the Pythagorean theorem:
a^2+b^2=c^2
a^2+2^2=7^2
a^2+4=49
a^2=45
a=â45
The answer is â45/7 or 3â5/7
2006-12-11 16:20:00 · answer #2 · answered by Anonymous · 1⤊ 0⤋
7^2 - (-2)^2 = 49 - 4 = 45
cos[sin-1(-2/7)] = (â45)/7
2006-12-11 15:51:11 · answer #3 · answered by Helmut 7 · 2⤊ 0⤋
Let x = arcsin (-2/7)
So sinx = -2/7
So x = 2Ï - y wher y is acute and siny = 2/7
so cos y = â(1 - (2/7)²)
= â(45/49)
=3â(5)/7
So cosx = cos (2Ï - y) = +cosy = 3â(5)/7
2006-12-11 15:46:01 · answer #4 · answered by Wal C 6 · 2⤊ 1⤋
sqrt(45)/7 is one answer
2006-12-11 15:42:59 · answer #5 · answered by shamu 2 · 2⤊ 0⤋
+/-(sqrt(45))/7
2006-12-11 17:35:41 · answer #6 · answered by Anonymous · 2⤊ 0⤋