Question

If sinx=1/9, x in quadrant 1, then find (without finding x):

Answer 1

Answer:

Part 1) $sin(2x)=8\frac{\sqrt{5}}{81}$

Part 2) $cos(2x)=\frac{79}{81}$

Part 3) $tan(2x)=8\frac{\sqrt{5}}{79}$

Step-by-step explanation:

Part 1) Find sin(2x)

we know that

$sin(2x)=2sin(x)cos(x)$

we have

$sin(x)=\frac{1}{9}$

Find cos(x)

Remember that

$sin^{2}(x)+cos^{2}(x)=1$

substitute

$(\frac{1}{9})^{2}+cos^{2}(x)=1$

$cos^{2}(x)=1-(\frac{1}{9})^{2}$

$cos^{2}(x)=1-(\frac{1}{81})$

$cos^{2}(x)=\frac{80}{81}$

$cos(x)=\frac{\sqrt{80}}{9}$

$cos(x)=4\frac{\sqrt{5}}{9}$ -----> is positive because angle x belong to the I quadrant

Find sin(2x)

we have

$sin(x)=\frac{1}{9}$

$cos(x)=4\frac{\sqrt{5}}{9}$

so

$sin(2x)=2sin(x)cos(x)$

$sin(2x)=2(\frac{1}{9})(4\frac{\sqrt{5}}{9})$

$sin(2x)=8\frac{\sqrt{5}}{81}$

Part 2) Find cos(2x)

we know that

$cos(2x)=cos^{2}(x)-sin^{2}(x)$

we have

$sin(x)=\frac{1}{9}$

$cos(x)=4\frac{\sqrt{5}}{9}$

so

$cos(2x)=(4\frac{\sqrt{5}}{9})^{2}-(\frac{1}{9})^{2}$

$cos(2x)=\frac{80}{81}-\frac{1}{81}$

$cos(2x)=\frac{79}{81}$

Part 3) Find tan(2x)

we know that

$tan(2x)=\frac{sin(2x)}{cos(2x)}$

we have

$sin(2x)=8\frac{\sqrt{5}}{81}$

$cos(2x)=\frac{79}{81}$

so

$tan(2x)=\frac{8\frac{\sqrt{5}}{81}}{\frac{79}{81}}$

$tan(2x)=8\frac{\sqrt{5}}{79}$