Let us consider LHS:

\[\mathbf{cos}\text{ }\mathbf{\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{2\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{4\pi }/\mathbf{15}\text{ }\mathbf{cos}\text{ }\mathbf{7\pi }/\mathbf{15}\text{ }\]

Let us multiply and divide by \[2\text{ }sin\text{ }\pi /15\], we get,

\[=\text{ }\left[ 2\text{ }sin\text{ }\pi /15\text{ }cos\text{ }\pi /15 \right]\text{ }cos\text{ }2\pi /15\text{ }cos\text{ }4\pi /15\text{ }cos\text{ }7\pi /15]\text{ }/\text{ }2\text{ }sin\text{ }\pi /15\]

We know, \[2sin\text{ }A\text{ }cos\text{ }A\text{ }=\text{ }sin\text{ }2A\]

Then the above equation becomes,

\[=\text{ }\left[ \left( sin\text{ }2\pi /15 \right)\text{ }cos\text{ }2\pi /15\text{ }cos\text{ }4\pi /15\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }2\text{ }sin\text{ }\pi /15\]

Now, multiply and divide by \[2\] we get,

\[=\text{ }\left[ \left( 2\text{ }sin\text{ }2\pi /15\text{ }cos\text{ }2\pi /15 \right)\text{ }cos\text{ }4\pi /15\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }2\text{ }\times \text{ }2\text{ }sin\text{ }\pi /15\]

We know, \[2sin\text{ }A\text{ }cos\text{ }A\text{ }=\text{ }sin\text{ }2A\]

Then the above equation becomes,

\[=\text{ }\left[ \left( sin\text{ }4\pi /15 \right)\text{ }cos\text{ }4\pi /15\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }4\text{ }sin\text{ }\pi /15\]

Now, multiply and divide by \[2\]we get,

\[=\text{ }\left[ \left( 2\text{ }sin\text{ }4\pi /15\text{ }cos\text{ }4\pi /15 \right)\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }2\text{ }\times \text{ }4\text{ }sin\text{ }\pi /15\]

We know, \[2sin\text{ }A\text{ }cos\text{ }A\text{ }=\text{ }sin\text{ }2A\]

Then the above equation becomes,

\[=\text{ }\left[ \left( sin\text{ }8\pi /15 \right)\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }8\text{ }sin\text{ }\pi /15\]

Now, multiply and divide by \[2\] we get,

\[=\text{ }\left[ 2\text{ }sin\text{ }8\pi /15\text{ }cos\text{ }7\pi /15 \right]\text{ }/\text{ }2\text{ }\times \text{ }8\text{ }sin\text{ }\pi /15\]

We know, \[2sin\text{ }A\text{ }cos\text{ }B\text{ }=\text{ }sin\text{ }\left( A+B \right)\text{ }+\text{ }sin\text{ }\left( AB \right)\]

Then the above equation becomes,

\[=\text{ }\left[ sin\text{ }\left( 8\pi /15\text{ }+\text{ }7\pi /15 \right)\text{ }+\text{ }sin\text{ }\left( 8\pi /15\text{ }\text{ }7\pi /15 \right) \right]\text{ }/\text{ }16\text{ }sin\text{ }\pi /15\]

\[=\text{ }\left[ sin\text{ }\left( \pi  \right)\text{ }+\text{ }sin\text{ }\left( \pi /15 \right) \right]\text{ }/\text{ }16\text{ }sin\text{ }\pi /15\]

\[=\text{ }\left[ 0\text{ }+\text{ }sin\text{ }\left( \pi /15 \right) \right]\text{ }/\text{ }16\text{ }sin\text{ }\pi /15\]

\[=\text{ }sin\text{ }\left( \pi /15 \right)\text{ }/\text{ }16\text{ }sin\text{ }\pi /15\]

\[=\text{ }1/16\]

= RHS

Hence proved.