Answer to Question #167794 in Civil and Environmental Engineering for eyel

Question

A.     Solve the following problems. Write your answer on a sheet of paper.

1.      Given that sin α =  and sin β = , find 𝑡𝑎𝑛 (α + β) if both α and β are in QIV.

2.      Given that csc A = , A in QI, and sec B = , sin B < 0, find

a.      cos (A – B)

b.     tan (A – B)

3.      If tan (x + y) =  and tan y = , what is tan x?

4.      The point (9, –5) lies on the terminal side of the angle 𝜃 in standard position. Find (sin 𝜃 + cos 𝜃).

B.      Solve the following problems. Write your answer on a sheet of paper.

1.      If cos t = , what is cos 2t?

2.      Use half – angle identities to find the exact value of tan 22.5⁰ and sin 15⁰.

Answer 1

1.

$$\sin \alpha=a, -1\leq a\leq0$$

$$\sin\beta=b, -1\leq a\leq0$$

$$\cos \alpha=\sqrt{1-\sin^2 \alpha}=\sqrt{1-a^2}$$

$$\cos\beta =\sqrt{1-\sin^2 \beta}=\sqrt{1-b^2}$$

$$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$$

$$=\dfrac{\dfrac{a}{\sqrt{1-a^2}}+\dfrac{b}{\sqrt{1-b^2}}}{1-\dfrac{a}{\sqrt{1-a^2}}\dfrac{b}{\sqrt{1-b^2}}}$$

$$=\dfrac{a\sqrt{1-b^2}+b\sqrt{1-a^2}}{\sqrt{1-a^2}\sqrt{1-b^2}-ab}$$

$$\tan(\alpha+\beta)=\dfrac{a\sqrt{1-b^2}+b\sqrt{1-a^2}}{\sqrt{1-a^2}\sqrt{1-b^2}-ab}$$

2.

$$\csc A=\dfrac{1}{\sin A}=a, a\geq 1$$

$$\sin A=\dfrac{1}{a},$$

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

$$\sec B=\dfrac{1}{\cos B}=b, |b|\geq 1$$

$$\cos B=\dfrac{1}{b}$$

$$\sin B=-\sqrt{1-\cos^2 B}=-\sqrt{1-(\dfrac{1}{b})^2}=-\dfrac{\sqrt{b^2-1}}{|b|}$$

a.

$$\cos(A-B)=\cos A\cos B+\sin A\sin B$$

$$=\dfrac{\sqrt{a^2-1}}{a}(\dfrac{1}{b})+\dfrac{1}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$$

$$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$$

$$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$$

b.

$$\tan(A-B)=\dfrac{\sin(A-B)}{\cos(a-B)}$$

$$\sin(A-B)=\sin A\cos B-\cos A\sin B$$

$$=\dfrac{1}{a}(\dfrac{1}{b})-\dfrac{\sqrt{a^2-1}}{a}(-\dfrac{\sqrt{b^2-1}}{|b|})$$

$$b\geq1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}-\sqrt{b^2-1}}{ab}$$

$$\sin(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{ab}$$

$$\tan(A-B)=\dfrac{1+\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}-\sqrt{b^2-1}}$$

$$b\leq-1, a\geq1:\cos(A-B)=\dfrac{\sqrt{a^2-1}+\sqrt{b^2-1}}{ab}$$

$$\sin(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{ab}$$

$$\tan(A-B)=\dfrac{1-\sqrt{(a^2-1)(b^2-1)}}{\sqrt{a^2-1}+\sqrt{b^2-1}}$$

3.

$$\tan(x+y)=\dfrac{\tan x+\tan y}{1-\tan x \tan y}$$

$$\tan (x+y)-\tan x \tan (x+y)\tan y=\tan x+\tan y$$

$$\tan x=\dfrac{\tan (x+y)-\tan y}{1+\tan (x+y) \tan y}$$

Given $\tan(x+y)=a, \tan y=b$

$$\tan x=\dfrac{a-b}{1+ab}$$

4.

$$\sqrt{(9)^2+(-5)^2}=\sqrt{106}$$

$$\sin \theta=\dfrac{-5}{\sqrt{106}}, \cos\theta=\dfrac{9}{\sqrt{106}}$$

$$\sin \theta+\cos \theta =-\dfrac{5}{\sqrt{106}}+\dfrac{9}{\sqrt{106}}=\dfrac{4}{\sqrt{106}}$$

$$\sin \theta+\cos \theta =\dfrac{4}{\sqrt{106}}$$

B.

1.     

$$\cos2t=2\cos^2t-1$$

Given $\cos t=a$

$$\cos2t=2a^2-1$$

2.

$$\sin 15\degree=\sqrt{\dfrac{1-\cos 30\degree}{2}}=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}$$

$$\sin 15\degree=\dfrac{\sqrt{2-\sqrt{3}}}{2}$$

$$\tan 22.5\degree=\sqrt{\dfrac{1-\cos 45\degree}{1+\cos45\degree}}=\sqrt{\dfrac{1-\dfrac{\sqrt{2}}{2}}{1+\dfrac{\sqrt{2}}{2}}}$$

$$=\sqrt{\dfrac{2-\sqrt{2}}{2+\sqrt{2}}}=\dfrac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1$$

$$\tan 22.5\degree=\sqrt{2}-1$$