Question

Use the information that sin⁡θ=35 and 90°<θ<180° to answer the following questions. Part A: Which equation shows cos2θ+sin2θ=1 correctly rewritten to isolate cos θ? Part B: What are the possible values of cos⁡ θ (if the quadrant is not considered)? Part C: What is the correct value of cos θ (based on the quadrant)?

Answer 1

Answer:

(a) $cos\theta = \sqrt{1-sin^2\theta}$

(b) $cos\theta = \frac{4}{5}$

(c) $cos\theta = -\frac{4}{5}$

Step-by-step explanation:

Given that  90°< θ<180°  and $sin\theta= \frac{3}{5}$

Since  90°< θ<180° so θ lies in second quadrant where sinθ is positive but cosθ is negative.

(a) The identity,

Sin²θ+cos²θ= 1

⇒cos²θ=1-sin²θ

$\Rightarrow cos\theta = \sqrt{1-sin^2\theta}$

(b)

$cos\theta = \sqrt{1-sin^2\theta}$    [ Since the quadrant is not considered]

$\Rightarrow cos\theta = \sqrt{1-(\frac{3}{5})^2 }$

$\Rightarrow cos\theta = \sqrt{1-(\frac{9}{25}) }$

$\Rightarrow cos\theta = \sqrt{(\frac{25-9}{25}) }$

$\Rightarrow cos\theta = \sqrt{\frac{16}{25} }$

$\Rightarrow cos\theta = \frac{4}{5}$

(c)

$cos\theta = -\sqrt{1-sin^2\theta}$    [ since cos θ is negative in second quadrant]

$\Rightarrow cos\theta = -\sqrt{1-(\frac{3}{5})^2 }$

$\Rightarrow cos\theta = -\sqrt{1-(\frac{9}{25}) }$

$\Rightarrow cos\theta = -\frac{4}{5}$