Question

Special Triangles help?

Evaluate the following, giving the exact value when possible:

a) sec(225°)
b) sin(-15°)
c) sin(75°)

Answer 1

The secant is 1/cos. 225 degrees lies in the third quadrant where cos is negative. cos(225)=cos(180+45)=-cos(45)=-1/sqrt(2). Hence sec(225)=1/cos(225)=1/(-1/sqrt(2))=-sqrt(2).

sin(-15)=sin(45-60)=sin(45)cos(60)-cos(45)sin(60)=(1/sqrt(2))*(1/2)-(1/sqrt(2))*(sqrt(3)/2). Therefore\[\sin(-15) = \frac{1-\sqrt{3}}{2\sqrt{2}}=\frac{\sqrt{2}-\sqrt{6}}{4}\]

Answer 2

sin(75) = sin(90-15)=sin(90)cos(15)-cos(90)sin(15)=1 x cos(15) - 0 x sin(15)

Now,
\[\cos(15)=\sqrt{1-\sin^2(15)}=\sqrt{1-(\frac{1-\sqrt{3}}{2\sqrt{2}})^2}\]\[=\sqrt{1-\frac{1-2\sqrt{3}+3}{4 \times 2}}=\sqrt{\frac{2+\sqrt{3}}{4}}\]

Answer 3

i.e. \[\sin(75)=\frac{\sqrt{2+\sqrt{3}}}{2}\]