Answer
384.6k+ views
Hint: In order to determine $\cos 67.5^\circ $ using half-angle identity, first we will determine the $\theta $. Then, choose the accurate half-angle identity and resolve the $ \pm $ sign, substitute the values from the trigonometric values from the trigonometric table, then by further evaluation, we will get the required value
Complete step by step solution:
Now, we need to determine $\cos 67.5^\circ $ using half-angle identity.
Therefore, the half-angle identity that we can use here is $1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}$
Let $\dfrac{\theta }{2} = 67.5^\circ $
Then, $\theta = 67.5^\circ \times 2$
$\theta = 135^\circ $
Now, substituting the values in the equation, we have,
$1 + \cos \left( {135^\circ } \right) = 2{\cos ^2}\dfrac{{\left( {135^\circ } \right)}}{2}$
Now, $135^\circ = 180^\circ - 45^\circ $
Thus, $2{\cos ^2}67.5^\circ = 1 + \cos \left( {135^\circ - 45^\circ } \right)$
$2{\cos ^2}67.5^\circ = 1 - \cos 45^\circ $
We know that from trigonometric ratios, $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
Therefore, $2{\cos ^2}67.5^\circ = 1 - \dfrac{1}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{2 - \sqrt 2 }}{2}$
${\cos ^2}67.5^\circ = \dfrac{{2 - \sqrt 2 }}{4}$
$\cos 67.5^\circ = \pm \sqrt {\dfrac{{2 - \sqrt 2 }}{4}} $
Hence,$\cos 67.5^\circ = \pm \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$
Note: Special cases of the sum and difference formulas for sine and cosine are known as the double-angle identities and the half-angle identities.
Let us know about the sum and difference formulas for sine and cosine. Three basic trigonometric identities involve the sums of angles. The functions involved in these identities are sine, cosine and tangent. We can use the angle sum identities to determine the function values of any angles. These identities are useful whenever expressions involving trigonometric functions need to be simplified.
The angle sum identities are
$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$
$\sin \left( {a + b} \right) = \cos a\sin b + \sin a\cos b$
$\sin \left( {a - b} \right) = \cos a\sin b - \sin a\cos b$
$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$
$\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$
Trigonometric table involves the relationship with the length and angles of the triangle. It is generally associated with the right-angled triangle, where one of the angles is always $90^\circ $.
The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself. The most common double angle formulas are,
$\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
$\cos 2\theta = 2{\cos ^2}\theta - 1$ .
Half-angle formulas allow the expression of trigonometric functions of angles equal to $\dfrac{\theta }{2}$ in terms of $\theta $, which can simplify the functions and make it easier to perform more complex calculations. The most commonly used half angle identities are,
$\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \theta }}{2}} $
$\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} $ .
Complete step by step solution:
Now, we need to determine $\cos 67.5^\circ $ using half-angle identity.
Therefore, the half-angle identity that we can use here is $1 + \cos \theta = 2{\cos ^2}\dfrac{\theta }{2}$
Let $\dfrac{\theta }{2} = 67.5^\circ $
Then, $\theta = 67.5^\circ \times 2$
$\theta = 135^\circ $
Now, substituting the values in the equation, we have,
$1 + \cos \left( {135^\circ } \right) = 2{\cos ^2}\dfrac{{\left( {135^\circ } \right)}}{2}$
Now, $135^\circ = 180^\circ - 45^\circ $
Thus, $2{\cos ^2}67.5^\circ = 1 + \cos \left( {135^\circ - 45^\circ } \right)$
$2{\cos ^2}67.5^\circ = 1 - \cos 45^\circ $
We know that from trigonometric ratios, $\cos 45^\circ = \dfrac{1}{{\sqrt 2 }}$
Therefore, $2{\cos ^2}67.5^\circ = 1 - \dfrac{1}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{\sqrt 2 - 1}}{{\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$
$2{\cos ^2}67.5^\circ = \dfrac{{2 - \sqrt 2 }}{2}$
${\cos ^2}67.5^\circ = \dfrac{{2 - \sqrt 2 }}{4}$
$\cos 67.5^\circ = \pm \sqrt {\dfrac{{2 - \sqrt 2 }}{4}} $
Hence,$\cos 67.5^\circ = \pm \dfrac{{\sqrt {2 - \sqrt 2 } }}{2}$
Note: Special cases of the sum and difference formulas for sine and cosine are known as the double-angle identities and the half-angle identities.
Let us know about the sum and difference formulas for sine and cosine. Three basic trigonometric identities involve the sums of angles. The functions involved in these identities are sine, cosine and tangent. We can use the angle sum identities to determine the function values of any angles. These identities are useful whenever expressions involving trigonometric functions need to be simplified.
The angle sum identities are
$\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b$
$\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b$
$\sin \left( {a + b} \right) = \cos a\sin b + \sin a\cos b$
$\sin \left( {a - b} \right) = \cos a\sin b - \sin a\cos b$
$\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$
$\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$
Trigonometric table involves the relationship with the length and angles of the triangle. It is generally associated with the right-angled triangle, where one of the angles is always $90^\circ $.
The trigonometric double angle formulas give a relationship between the basic trigonometric functions applied to twice an angle in terms of trigonometric functions of the angle itself. The most common double angle formulas are,
$\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta $
$\cos 2\theta = 1 - 2{\sin ^2}\theta $
$\cos 2\theta = 2{\cos ^2}\theta - 1$ .
Half-angle formulas allow the expression of trigonometric functions of angles equal to $\dfrac{\theta }{2}$ in terms of $\theta $, which can simplify the functions and make it easier to perform more complex calculations. The most commonly used half angle identities are,
$\cos \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 + \cos \theta }}{2}} $
$\sin \left( {\dfrac{\theta }{2}} \right) = \pm \sqrt {\dfrac{{1 - \cos \theta }}{2}} $ .
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)