
How do you verify $\cos ({180^ \circ } - \theta ) = - \cos \theta $?
Answer
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Hint: We can solve this question by applying the trigonometric formula of $\cos (a - b) = \cos (a)\cos (b) + \sin (a)\sin (b)$, where a and b are angles. In the fourth quadrant the cosine remains positive along with the first quadrant where all the wanted sine, cosine, tangent remains positive. Also remember the trigonometry table so that we can quickly assign the values as per the angles . For example, $\cos {180^ \circ }$has the value assigned as -1 in the trigonometric table . Also $\sin {180^ \circ }$is equal to zero.
Complete step-by-step solution:
In order to verify that $\cos ({180^ \circ } - \theta ) = - \cos \theta $, we are going to apply the trigonometry formula $\cos (a - b) = \cos (a)\cos (b) + \sin (a)\sin (b)$.
In order to verify we will put $a = {180^ \circ }$and $b = \theta $ and substitute in the trigonometric formula.
On applying we get,
$\Rightarrow \cos ({180^ \circ } - \theta ) = \cos ({180^ \circ })\cos (\theta ) + \sin ({180^ \circ })\sin (\theta )$
Since, $\cos {180^ \circ } = - 1$ and $\sin {180^ \circ } = 0$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - 1 \times \cos (\theta ) + 0 \times \sin (\theta )$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - 1 \times \cos (\theta ) + 0$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - \cos \theta $
Hence, verified.
Hence the answer is $ \cos ({180^ \circ } - \theta ) = - \cos \theta $
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
It should be very clear that \[sin\left( {A + B} \right)\]is not equal to $\sin A + \sin B$.
No sine or cosine can ever be more than 1 as the ratio has the hypotenuse as its denominator.
Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.
Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
Complete step-by-step solution:
In order to verify that $\cos ({180^ \circ } - \theta ) = - \cos \theta $, we are going to apply the trigonometry formula $\cos (a - b) = \cos (a)\cos (b) + \sin (a)\sin (b)$.
In order to verify we will put $a = {180^ \circ }$and $b = \theta $ and substitute in the trigonometric formula.
On applying we get,
$\Rightarrow \cos ({180^ \circ } - \theta ) = \cos ({180^ \circ })\cos (\theta ) + \sin ({180^ \circ })\sin (\theta )$
Since, $\cos {180^ \circ } = - 1$ and $\sin {180^ \circ } = 0$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - 1 \times \cos (\theta ) + 0 \times \sin (\theta )$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - 1 \times \cos (\theta ) + 0$
$\Rightarrow \cos ({180^ \circ } - \theta ) = - \cos \theta $
Hence, verified.
Hence the answer is $ \cos ({180^ \circ } - \theta ) = - \cos \theta $
Note: Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.
It should be very clear that \[sin\left( {A + B} \right)\]is not equal to $\sin A + \sin B$.
No sine or cosine can ever be more than 1 as the ratio has the hypotenuse as its denominator.
Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.
Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
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