How do you find the exact functional value $cos\, 15^\circ$ using the cosine sum or difference identity?
Answer
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Hint: When we use any trigonometric sum or difference identity, we must assume only that angles ($a$ and $b$) in the identity, of whose trigonometric ratios are known to us.
Complete step by step solution:
As we have to find the value of $\cos {15^o}$, we will use the cosine difference identity, i.e.
$\cos (a - b) = \cos a\cos b + \sin a\sin b$
Let $a = {45^o}$ and $b = {30^o}$ as the value of $\cos a$, $\cos b$, $\sin a$ and $\sin b$ are known to us. We will substitute the value of and in the cosine difference identity so that it becomes,
$
\Rightarrow \cos (a - b) = \cos a\cos b + \sin a\sin b \\
\Rightarrow \cos ({45^o} - {30^o}) = \cos {45^o}\cos {30^o} + \sin {45^o}\sin {30^o} \\
\\
$
We have,
$\cos {45^o} = \dfrac{1}{{\sqrt 2 }}$; $\sin {45^o} = \dfrac{1}{{\sqrt 2 }}$; $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$; $\sin {30^o} = \dfrac{1}{2}$
On using these values, we get
$ \Rightarrow \cos {15^o} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
On multiplying the values, we get
$ \Rightarrow \cos {15^o} = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}$
Taking $2\sqrt 2 $ as the denominator and further simplifying, we get
$ \Rightarrow \cos {15^o} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Substituting $\sqrt 3 = 1.732$ and $\sqrt 2 = 1.414$ above, we have
$ \Rightarrow \cos {15^o} = \dfrac{{1.732 + 1}}{{2 \times 1.414}}$
$ \Rightarrow \cos {15^o} = \dfrac{{2.732}}{{2.828}}$
$ \Rightarrow \cos {15^o} = 0.966$
Hence, the exact functional value of $\cos {15^o}$ found using cosine difference identity is $0.966$.
Note: In case we don’t know the cosine sum or difference identity, we can use the sine sum or difference identity. But for this we will have to convert cosine angle into sine angle using complementary angles formulae, i.e. $\cos \theta = \sin (90 - \theta )$. In the given question, $\cos {15^o} = \sin ({90^o} - {15^o}) = \sin {75^o}$. We can now use the sine sum identity $\sin (a + b) = \sin a\cos b + \sin b\cos a$ to find the value of $\sin {75^o}$, where $a = {45^o}$ and $b = {30^o}$.
Complete step by step solution:
As we have to find the value of $\cos {15^o}$, we will use the cosine difference identity, i.e.
$\cos (a - b) = \cos a\cos b + \sin a\sin b$
Let $a = {45^o}$ and $b = {30^o}$ as the value of $\cos a$, $\cos b$, $\sin a$ and $\sin b$ are known to us. We will substitute the value of and in the cosine difference identity so that it becomes,
$
\Rightarrow \cos (a - b) = \cos a\cos b + \sin a\sin b \\
\Rightarrow \cos ({45^o} - {30^o}) = \cos {45^o}\cos {30^o} + \sin {45^o}\sin {30^o} \\
\\
$
We have,
$\cos {45^o} = \dfrac{1}{{\sqrt 2 }}$; $\sin {45^o} = \dfrac{1}{{\sqrt 2 }}$; $\cos {30^o} = \dfrac{{\sqrt 3 }}{2}$; $\sin {30^o} = \dfrac{1}{2}$
On using these values, we get
$ \Rightarrow \cos {15^o} = \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$
On multiplying the values, we get
$ \Rightarrow \cos {15^o} = \dfrac{{\sqrt 3 }}{{2\sqrt 2 }} + \dfrac{1}{{2\sqrt 2 }}$
Taking $2\sqrt 2 $ as the denominator and further simplifying, we get
$ \Rightarrow \cos {15^o} = \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$
Substituting $\sqrt 3 = 1.732$ and $\sqrt 2 = 1.414$ above, we have
$ \Rightarrow \cos {15^o} = \dfrac{{1.732 + 1}}{{2 \times 1.414}}$
$ \Rightarrow \cos {15^o} = \dfrac{{2.732}}{{2.828}}$
$ \Rightarrow \cos {15^o} = 0.966$
Hence, the exact functional value of $\cos {15^o}$ found using cosine difference identity is $0.966$.
Note: In case we don’t know the cosine sum or difference identity, we can use the sine sum or difference identity. But for this we will have to convert cosine angle into sine angle using complementary angles formulae, i.e. $\cos \theta = \sin (90 - \theta )$. In the given question, $\cos {15^o} = \sin ({90^o} - {15^o}) = \sin {75^o}$. We can now use the sine sum identity $\sin (a + b) = \sin a\cos b + \sin b\cos a$ to find the value of $\sin {75^o}$, where $a = {45^o}$ and $b = {30^o}$.
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