Fundamental Trigonometric Identities

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1. CSCA Math December 2025 Past Paper
   $\left(\sin\frac{\pi}{4}+\cos\frac{\pi}{4}\right)\tan\frac{\pi}{4}=$ ( ).
   • A．$2\sqrt{2}$
   • B．1
   • C．$\sqrt{2}$
   • D．2
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   Jawaban: C

   Answer: C $\sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$ and $\tan\frac{\pi}{4}=1$, so the value is $\sqrt{2}$. Choose C.
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2. CSCA Math January 2026 Past Paper
   Given that $\sin \alpha = \frac { 4 } { 5 }$, and the angle $\alpha$ belongs to the second quadrant, then which of the following statements is correct? ()
   • A．$\cos \alpha = \frac { 3 } { 5 }$
   • B．$\cos \alpha = - \frac { 3 } { 5 }$
   • C．$\tan \alpha = - \frac { 3 } { 4 }$
   • D．$\tan \alpha = \frac { 4 } { 3 }$
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   Jawaban: B

   Given $\sin \alpha = 4 / 5$ and $\alpha$ is in Quadrant II. Find $\cos \alpha$. In Quadrant II, sine is positive and cosine is negative. Use $\sin ^ { 2 } \alpha + \cos ^ { 2 } \alpha = 1 : \cos ^ { 2 } \alpha = 1 - ( 4 / 5 ) ^ { 2 } = 1 - 16 / 25 = 9 / 25$. So $\cos \alpha = - 3 / 5$.
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3. CSCA Math April 2026 Past Paper
   If point $P ( x , y )$ lies on the terminal side of angle $\alpha$ with $r = \sqrt { x ^ { 2 } + y ^ { 2 } }$, then $\cos \alpha =$
   • A．$\frac { y } { r }$
   • B．$\frac { y } { x }$
   • C．$\frac { x } { y }$
   • D．$\frac { x } { r }$
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   Jawaban: D

   By definition $\cos \alpha = \frac { x } { r }$. (Note: $\frac { y } { r } = \sin \alpha , \frac { y } { x } = \tan \alpha$.)
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