Calculate sin(x) (or cos(x)) and tan(x) when cos(x) (or sin(x)) is known Algebra II

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\cos\left(x\right)=0.6

Therefore:

sin^2\left(x\right)+\left(.6\right)^2=1

sin^2\left(x\right)+.36=1

sin^2\left(x\right)=.64

Therefore:

\sin\left(x\right)=\pm .8

It is assumed that \sin\left(x\right) \gt 0, hence:

\sin\left(x\right)= .8

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)=\dfrac{.8}{.6}=\dfrac{4}{3}

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\sin\left(x\right)=0.3

Therefore:

cos^2\left(x\right)+\left(.3\right)^2=1

cos^2\left(x\right)+.09=1

cos^2\left(x\right)=.91

Therefore:

\cos\left(x\right)\approx \pm .95

It is assumed that \cos\left(x\right) \gt 0, hence:

\cos\left(x\right)\approx .95

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)=\dfrac{.3}{.95}\approx.31

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\sin\left(x\right)=0.6

Therefore:

cos^2\left(x\right)+.36=1

cos^2\left(x\right)=.64

Therefore:

\cos\left(x\right)=\pm .8

It is assumed that \cos\left(x\right) \leq 0, hence:

\cos\left(x\right)= -.8

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)=\dfrac{.6}{-.8}=\dfrac{-3}{4}

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\sin\left(x\right)=0.2

Thus:

cos^2\left(x\right)+.04=1

cos^2\left(x\right)=.96

And:

\cos\left(x\right)\approx \pm .98

It is assumed that \cos\left(x\right) \geq 0, hence:

\cos\left(x\right)\approx .98

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)\approx \dfrac{.2}{.98}\approx.20

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\sin\left(x\right)=0.5

Thus:

cos^2\left(x\right)+.25=1

cos^2\left(x\right)=.75

And:

\cos\left(x\right)\approx \pm .87

It is assumed that \cos\left(x\right) \geq 0, hence:

\cos\left(x\right)\approx .87

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)\approx \dfrac{.5}{.87}\approx.57

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\cos\left(x\right)=0.3

Thus:

sin^2\left(x\right)+\left(.3\right)^2=1

sin^2\left(x\right)+.09=1

sin^2\left(x\right)=.91

And:

\sin\left(x\right)\approx \pm .95

It is assumed that \sin\left(x\right) \leq 0, hence:

\sin\left(x\right)\approx -.95

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)=\dfrac{-.95}{.3}\approx -3.16

For any real number x:

sin^2\left(x\right)+cos^2\left(x\right)=1

Here we know that:

\cos\left(x\right)=0.2

Thus:

sin^2\left(x\right)+\left(.2\right)^2=1

sin^2\left(x\right)+.04=1

sin^2\left(x\right)=.96

And:

\sin\left(x\right)\approx \pm .98

It is assumed that \sin\left(x\right) \geq 0, hence:

\sin\left(x\right)\approx .98

Furthermore for any real number x such that \cos\left(x\right)\neq0 :

\tan\left(x\right)=\dfrac{\sin\left(x\right)}{\cos\left(x\right)}

Therefore:

\tan\left(x\right)\approx \dfrac{.98}{.2}\approx 4.9