Respuesta :

Solution:

From the sum and difference identities expressed as

[tex]\begin{gathered} \sin (A+B)=\sin A\cos B+\sin B\cos A\text{ ---- equation 1} \\ \sin (A-B)=\sin A\cos B-\sin B\cos A\text{ ----- equation 2} \\ \text{cos}(A+B)=\cos A\cos B-\sin A\sin B\text{ ------ equation 3} \\ \text{cos}(A-B)=\cos A\cos B+\sin A\sin B\text{ ------ equation }4 \end{gathered}[/tex]

Given:

[tex]\cos (15)\cos (15)-\sin (15)\sin (15)[/tex]

The above expression satisfies equation 3, where

[tex]\begin{gathered} A=15, \\ B=15 \end{gathered}[/tex]

Thus,

[tex]\begin{gathered} \cos (15)\cos (15)-\sin (15)\sin (15)=\cos (15+15) \\ =\cos (30) \end{gathered}[/tex]

Hence, the expression is simplified to be

[tex]\cos (30)[/tex]

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