We have an answer to "How do you find exact value of tan⁡(π4)?"

High School
Geometry
Trigonometry
Trigonometric equations and identities

Haven

Answered question

2021-10-25

How do you find exact value of

\tan (\frac{π}{4})

Answer & Explanation

Macsen Nixon

Skilled2021-10-26Added 117 answers

Given:

\tan (\frac{π}{4})

The cousine is defined as the opposite leg divided by the hypotenuse of a rectangular triangle, while the sine is defined as the adjacent leg divided by the hypotenuse.

\sin (\frac{π}{4}) = \frac{adjacent leg}{hypotenuse}

\cos (\frac{π}{4}) = \frac{opposite leg}{hypotenuse}

By the Pythagorean theorem

a^{2} + b^{2} = c^{2}

where a (adjacent) and b (opposite) are the legs and c is the hypotenuse c. This then implies that

c = \sqrt{a^{2} + b^{2}}

for the hypotenuse c.

\sin (\frac{π}{4}) = \frac{a}{\sqrt{a^{2} + b^{2}}}

\cos (\frac{π}{4}) = \frac{b}{\sqrt{a^{2} + b^{2}}}

However, if one of the angles of the triangle is

45^{\circ}

\frac{π}{4}

, then the triangle is isosceles and thus the two legs have the same length a=b

\cos (\frac{π}{4}) = \frac{b}{\sqrt{a^{2} + b^{2}}}

= \frac{b}{\sqrt{b^{2} + b^{2}}}

= \frac{b}{\sqrt{2 b^{2}}}

= \frac{b}{\sqrt{2} \sqrt{b^{2}}}

= \frac{b}{\sqrt{2} b}

= \frac{1}{\sqrt{2}}

= \frac{1 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}}

= \frac{\sqrt{2}}{{(\sqrt{2})}^{2}}

= \frac{\sqrt{2}}{2}

\sin (\frac{π}{4}) = \frac{a}{\sqrt{a^{2} + b^{2}}} = \frac{b}{\sqrt{a^{2} + b^{2}}} = \cos (\frac{π}{4})

The tangent is the sine divided by the cousine:

\tan (\frac{π}{4}) = \frac{\sin (\frac{π}{4})}{\cos (\frac{π}{4})} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = 1

Result: 1

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Trigonometry

Didn't find what you were looking for?