Question

Derive the value of sin45 degree, cos45 degree and tan45 degree

Answer 1

$\large\underline{\sf{Solution-}}$

Let us consider a right angle isosceles triangle ABC right - angled at B such that AB = BC.

Since, AB = BC = x (say)

We know, angle opposite to equal sides are equal.

⇛ ∠ACB = ∠BAC = y ( say ).

Now, we know

Sum of all interior angles of a triangle is supplementary.

⇛ ∠ABC + ∠BAC + ∠BCA = 180°

⇛ 90° + y + y = 180°

⇛ 90° + 2y = 180°

⇛ 2y = 180° - 90°

⇛ 2y = 90°

⇛ y = 45°

Now, In right triangle ABC,

Using Pythagoras Theorem, we have

$\red{\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2} \: }$

$\red{\rm :\longmapsto\: {AC}^{2} = {x}^{2} + {x}^{2} \: }$

$\red{\rm :\longmapsto\: {AC}^{2} = 2{x}^{2} \: }$

$\red{\rm \implies\:AC = \sqrt{2}x}$

Now, Consider,

$\red{\rm :\longmapsto\:sin45\degree = \dfrac{AB}{AC} }$

$\red{\rm :\longmapsto\:sin45\degree = \dfrac{x}{ \sqrt{2} \: x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: sin45\degree = \frac{1}{ \sqrt{2} } \: }}}$

Now, Consider

$\red{\rm :\longmapsto\:cos45\degree = \dfrac{BC}{AC} }$

$\red{\rm :\longmapsto\:cos45\degree = \dfrac{x}{ \sqrt{2} \: x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: cos45\degree = \frac{1}{ \sqrt{2} } \: }}}$

Now, Consider

$\red{\rm :\longmapsto\:tan45\degree = \dfrac{AB}{BC} }$

$\red{\rm :\longmapsto\:tan45\degree = \dfrac{x}{x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: \:tan45\degree = 1 }}}$

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Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$

Answer 2

$\large\underline{\sf{Solution-}}$

Let us consider a right angle isosceles triangle ABC right - angled at B such that AB = BC.

Since, AB = BC = x (say)

We know, angle opposite to equal sides are equal.

⇛ ∠ACB = ∠BAC = y ( say ).

Now, we know

Sum of all interior angles of a triangle is supplementary.

⇛ ∠ABC + ∠BAC + ∠BCA = 180°

⇛ 90° + y + y = 180°

⇛ 90° + 2y = 180°

⇛ 2y = 180° - 90°

⇛ 2y = 90°

⇛ y = 45°

Now, In right triangle ABC,

Using Pythagoras Theorem, we have

$\red{\rm :\longmapsto\: {AC}^{2} = {AB}^{2} + {BC}^{2} \: }$

$\red{\rm :\longmapsto\: {AC}^{2} = {x}^{2} + {x}^{2} \: }$

$\red{\rm :\longmapsto\: {AC}^{2} = 2{x}^{2} \: }$

$\red{\rm \implies\:AC = \sqrt{2}x}$

Now, Consider,

$\red{\rm :\longmapsto\:sin45\degree = \dfrac{AB}{AC} }$

$\red{\rm :\longmapsto\:sin45\degree = \dfrac{x}{ \sqrt{2} \: x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: sin45\degree = \frac{1}{ \sqrt{2} } \: }}}$

Now, Consider

$\red{\rm :\longmapsto\:cos45\degree = \dfrac{BC}{AC} }$

$\red{\rm :\longmapsto\:cos45\degree = \dfrac{x}{ \sqrt{2} \: x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: cos45\degree = \frac{1}{ \sqrt{2} } \: }}}$

Now, Consider

$\red{\rm :\longmapsto\:tan45\degree = \dfrac{AB}{BC} }$

$\red{\rm :\longmapsto\:tan45\degree = \dfrac{x}{x} }$

$\red{\rm \implies\:\boxed{ \tt{ \: \:tan45\degree = 1 }}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Additional Information :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$