
How do you use the half angle formula to find \[\sin {75^ \circ }\]?
Answer
563.1k+ views
Hint: Half-angle formulas allow the expression of trigonometric functions of angles equal to \[\dfrac{\alpha }{2}\] in terms of \[\alpha \], which can simplify the functions and make it easier to perform more complex calculations, such as integration, on them. Half-angle formulas are especially useful in finding the values of unknown trigonometric functions.
Formula used: To find the value of \[\sin \theta \] by using half angle we will use the formula of \[\sin \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{2}} \].
Complete step-by-step solution:
We have to find the value of \[\sin {75^ \circ }\] by using the half angle formula.
We know that,
\[\sin \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{2}} \] is the half angle formula of \[\sin \theta \].
Substitute \[\theta = {75^ \circ }\] we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos 2 \times {{75}^ \circ }}}{2}} \]
On multiply the terms and we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos {{150}^ \circ }}}{2}} \]
On splitting we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos ({{180}^ \circ } - {{30}^ \circ })}}{2}} \]
Simplifying again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 + \cos 30}}{2}} \]
On putting the values and again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{2}} \]
On simplifying again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}} \]
Hence, the value of \[\sin {75^ \circ } = \sqrt {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}} \]
Note: Cutting special angles on the unit circle in half gives you a variety of new angles that can’t be achieved by using the sum and difference formulas or the double-angle formulas. Although the half-angle formulas won’t give you all the angles of the unit circle, they certainly get you closer than you were before.
We also can find the values of trig functions for angles like pi/12, each of which are exactly half of angles on the unit circle. These angles aren’t the only types that the identities work for. we can continue to halve any angle on the unit circle and take the trig functions of those angles. For example, 15 degrees is half of 30 degrees, and 7.5 degrees is half of 15 degrees.
In the half-angle formula for sine and cosine, notice that a plus/minus sign appears in front of each radical (square root). Whether the answer is positive or negative depends on which quadrant the new angle (the half angle) is in. The half-angle formula for tangent doesn’t have a plus/minus sign in front, so the above doesn’t apply to tangent.
Formula used: To find the value of \[\sin \theta \] by using half angle we will use the formula of \[\sin \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{2}} \].
Complete step-by-step solution:
We have to find the value of \[\sin {75^ \circ }\] by using the half angle formula.
We know that,
\[\sin \theta = \sqrt {\dfrac{{1 - \cos 2\theta }}{2}} \] is the half angle formula of \[\sin \theta \].
Substitute \[\theta = {75^ \circ }\] we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos 2 \times {{75}^ \circ }}}{2}} \]
On multiply the terms and we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos {{150}^ \circ }}}{2}} \]
On splitting we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 - \cos ({{180}^ \circ } - {{30}^ \circ })}}{2}} \]
Simplifying again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 + \cos 30}}{2}} \]
On putting the values and again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{{1 + \dfrac{{\sqrt 3 }}{2}}}{2}} \]
On simplifying again we get,
\[ \Rightarrow \sin {75^ \circ } = \sqrt {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}} \]
Hence, the value of \[\sin {75^ \circ } = \sqrt {\dfrac{1}{2} + \dfrac{{\sqrt 3 }}{4}} \]
Note: Cutting special angles on the unit circle in half gives you a variety of new angles that can’t be achieved by using the sum and difference formulas or the double-angle formulas. Although the half-angle formulas won’t give you all the angles of the unit circle, they certainly get you closer than you were before.
We also can find the values of trig functions for angles like pi/12, each of which are exactly half of angles on the unit circle. These angles aren’t the only types that the identities work for. we can continue to halve any angle on the unit circle and take the trig functions of those angles. For example, 15 degrees is half of 30 degrees, and 7.5 degrees is half of 15 degrees.
In the half-angle formula for sine and cosine, notice that a plus/minus sign appears in front of each radical (square root). Whether the answer is positive or negative depends on which quadrant the new angle (the half angle) is in. The half-angle formula for tangent doesn’t have a plus/minus sign in front, so the above doesn’t apply to tangent.
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