Question

Write an expression using all six trigonometric functions such that the value of the expression is 3.

Answer 1

The six trigonometric functions are sin, cos, tan, csc, sec and ctg.

And you know these basic relations:

sin^2 (x) + cos^2 (x) = 1

tan(x)* ctg(x) = 1

csc(x)* sin(x) = 1

sec(x)*cos(x) = 1

so, you can write:

sin^2 (x) + cos^2 (x) + csc(x)*sin(x) + sec(x)*cos(x) + tan(x)*ctg(x) - 1 = 3

Which you can verify because:

1 + 1 + 1 + 1 - 1 = 3.

Answer 2

To write the expression using all six trigonometric function, we need to understand about trigonometric function relation.

The trigonometric expression are as follows,

$\sin^2 (x) + \cos^2 (x) + \csc(x)\times \sin(x) + \sec(x)\times \cos(x) + \tan(x)\times \cot (x) - 1$

Given:

The trigonometric function are as follows,

$\sin, \cos, \tan, \csc, \sec \:\rm and \:\cot.$

Write the relationship between $\sin$ and $\cos$.

$\sin^2 (x) + \cos^2 (x) = 1$

Write the relationship between $\tan$ and $\cot$.

$\tan(x)\times \cot(x) = 1$

Write the relationship between $\csc$ and $\sin$.

$\csc(x)\times \sin(x) = 1$

Write the relationship between $\sec$ and $\cos$.

$\sec(x)*\cos(x) = 1$

Now combine all the above relationship

$\sin^2 (x) + \cos^2 (x) + \csc(x)\times sin(x) + \sec(x)\times \cos(x) +\tan(x)\times \cot(x) - 1 = 3$

Now solve further,

$1+1+1+1-1=3\\3=3$

Thus, the trigonometric expression are as follows,

$\sin^2 (x) + \cos^2 (x) + \csc(x)\times \sin(x) + \sec(x)\times \cos(x) + \tan(x)\times \cot (x) - 1$

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