Solve the equation sine[Arccos 2/7]=

Question

Please show step by step solution and explanation.

Answer 1

Picture a right triangle where x = 2 and r = 7.

Now use Pythagorean Theorem to find y.

x^2 + y^2 = r^2

2^2 + y^2 = 7^2

Y^2 = 49 - 4 = 45

y = Sqrt45 = 3Sqrt5

Sin[Arccos(2/7)] = (3Sqrt5)/7

Answer 2

Hi Saja!

Range of Arccos occurs in quadrants 1 and 2.

Since 2/7 is positive, this must occur in quadrant 1 for cosine.

Let θ = arccos (2/7), then cos θ = 2/7.

y

| /|

| 7 / |

| / | b

| / θ |

---------------------------- x

2

Use the pythagorean theorem to find b.

a² + b² = c²

2² + b² = 7²

4 + b² = 49

-4 -4 subtract 4 from both sides

----------------

b² = 45

b = √45 take square root of both sides

b = √(9*5)

b = 3√5

opp 3√5

sin θ = ---- = ------

hyp 7

This is also sin (arccos (2/7) ) since we let θ = arccos (2/7)