You need the Unit Circle to answer these questions! (attached)
7. Answer: [tex]\bold{\dfrac{\sqrt6-\sqrt2}{4}}[/tex]
Step-by-step explanation:
[tex]sin\dfrac{\pi}{12}=15^o\\\\\text{Since }15^o\text{ is not on the Unit Circle, you will need to use the difference}\\\text{formula for sin: sin(A - B) = sinAcosB - cosAsinB}\\\\sin15=sin(45-30)=\dfrac{\sqrt2}{2}\cdot\dfrac{\sqrt3}{2}-\dfrac{\sqrt2}{2}\cdot\dfrac{1}{2}=\dfrac{\sqrt6}{4}-\dfrac{\sqrt2}{4}=\boxed{\dfrac{\sqrt6-\sqrt2}{4}}[/tex]
8. Answer: 0
Step-by-step explanation:
The sum formula for cos is: cos (A + B) = cosA cosB - sinA sinB
Since we are given cos75° cos15° - sin75° sin15°, we can conclude that we are looking for cos (75° + 15°) = cos 90°.
The Unit Circle shows us that cos 90° = 0
9. Answer: [tex]\bold{\dfrac{\sqrt3}{3}}[/tex]
Step-by-step explanation:
The given expression is for tan (A + B).
tan (10° + 20°) = tan 30° = [tex]\dfrac{sin30^o}{cos30^o}=\dfrac{1}{\sqrt3}\implies\boxed{\dfrac{\sqrt3}{3}}[/tex]
10. Answer: [tex]\bold{\pm\dfrac{2-\sqrt2}{2}}[/tex]
Step-by-step explanation:
[tex]sin\dfrac{\pi}{8}=22.5^o=sin\dfrac{45^o}{2}\\\text{So we need to use the half-angle formula for sin:}\\\\sin\dfrac{\theta}{2}=\pm\sqrt\dfrac{1-cos\theta}{2}}\\\\sin\dfrac{45^o}{2}=\pm\sqrt\dfrac{1-cos45^o}{2}}=\pm\sqrt\dfrac{1-\frac{\sqrt2}{2}}{2}}=\pm\sqrt\dfrac{\frac{2}{2}-\frac{\sqrt2}{2}}{\frac{4}{2}}}=\boxed{\pm\dfrac{2-\sqrt2}{2}}[/tex]
11. TOO BLURRY TO SEE THE PROBLEM
12. Answer: [tex]\bold{\dfrac{7}{25}}[/tex]
Step-by-step explanation:
Quadrant II means that sin is positive and cos is negative.
[tex]sin\ x=\dfrac{3}{5}\implies cos\ x=-\dfrac{4}{5}\\\\cos\ 2x=cos^2\ x-sin^2\ x\\\\.\qquad=\bigg(\dfrac{-4}{5}\bigg)^2-\bigg(\dfrac{3}{5}\bigg)^2\\\\.\qquad=\dfrac{16}{25}-\dfrac{9}{25}\\\\.\qquad=\boxed{\dfrac{7}{25}}[/tex]
