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InterviewSolution
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निम्नलिखित व्यंजकों के मान ज्ञात कीजिए - `sin^(-1)(sin.(3pi)/(5))` |
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Answer» चूँकि `sin^(-1)` की मुख्य शाखा का परिसर `[-(pi)/(2),(pi)/(2)]` होता है। `therefore" "sin^(-1)(sin.(3pi)/(5))ne (3pi)/(5)," "[because (3pi)/(5) cancel(in)[-(pi)/(2),(pi)/(2)]]` अब , `sin^(-1)(sin.(3pi)/(5))` `=sin^(-1)[sin(pi-(2pi)/(5))]," "[because (3pi)/(5)=pi-(25)/(5)]` `=sin^(-1)(sin.(2pi)/(5))," "[because sin(pi-theta)=sin theta]` `=(2pi)/(5)in[-(pi)/(2),(pi)/(3)]," "[because sin^(-1)(sin theta)=theta]` `therefore" "sin^(-1)(sin.(3pi)/(5))=(2pi)/(5).` |
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