• 2021 April(Z04) Math Question 42

    Trigonometry

    - Pythagorean Theorem & sin cos tan

    Pythagorean Theorem and Special Right Triangles
    Choice J
    (sin60o)(cos30o)(sin30o)(cos60o)=?(sin60^o)(cos30^o)(sin30^o)(cos60^o)=?

    As we know that

    sin60^o=\frac\{\sqrt3\}\{2\},\quad cos30^o=\frac\{\sqrt3\}\{2\},\quad sin30^o=\frac\{1\}\{2\},\quad cos60^o=\frac\{1\}\{2\}

    So, let's plug the values into the given equation

    (sin60^o)(cos30^o)(sin30^o)(cos60^o)=\left(\frac\{\sqrt3\}\{2\}\right)\left(\frac\{\sqrt3\}\{2\}\right)+\left(\frac\{1\}\{2\}\right)\left(\frac\{1\}\{2\}\right)
    ={3}{4}+{1}{4}=\frac\{3\}\{4\}+\frac\{1\}\{4\}
    ={3+1}{4}=\frac\{3+1\}\{4\}
    =1=1

    Also, we can see that

    sin(60o+30o)=sin90o=1sin(60^o+30^o)=sin90^o=1

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