اتحادهای مثلثاتی (Trigonometric Identities) (اتحادهای توابع مثلثاتی)، در ریاضیات (Mathematics)
\[ \eqalign{
& \sin ( - \theta ) = - \sin \theta \cr
& \cos ( - \theta ) = \cos \theta \cr}
\]
\[ {\sin ^2}\theta + {\cos ^2}\theta = 1
\]
\[ \eqalign{
& {\sec ^2}\theta = 1 + {\tan ^2}\theta \cr
& {\csc ^2}\theta = 1 + {\cot ^2}\theta \cr}
\]
\[ \eqalign{
& \sin 2\theta = 2\sin \theta \cos \theta \cr
& \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta \cr}
\]
\[ \cos 2\theta = {\cos ^2}\theta - {\sin ^2}\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta
\]
\[ \eqalign{
& {\cos ^2}\theta = {{1 + \cos 2\theta } \over 2} \cr
& {\sin ^2}\theta = {{1 - \cos 2\theta } \over 2} \cr}
\]
\[ \eqalign{
& \sin (A + B) = \sin A\cos B + \cos A\sin B \cr
& \sin (A - B) = \sin A\cos B - \cos A\sin B \cr
& \cos (A + B) = \cos A\cos B - \sin A\sin B \cr
& \cos (A - B) = \cos A\cos B + \sin A\sin B \cr}
\]
\[ \eqalign{
& \tan (A + B) = {{\tan A + \tan B} \over {1 - \tan A\tan B}} \cr
& \tan (A - B) = {{\tan A - \tan B} \over {1 + \tan A\tan B}} \cr}
\]
\[ \eqalign{
& \sin (A - {\pi \over 2}) = - \cos A \cr
& \cos (A - {\pi \over 2}) = \sin A \cr
& \sin (A + {\pi \over 2}) = \cos A \cr
& \cos (A + {\pi \over 2}) = - \sin A \cr}
\]
\[ \eqalign{
& \sin A\sin B = {1 \over 2}\cos (A - B) - {1 \over 2}\cos (A + B) \cr
& \cos A\cos B = {1 \over 2}\cos (A - B) + {1 \over 2}\cos (A + B) \cr
& \sin A\cos B = {1 \over 2}\sin (A - B) + {1 \over 2}\sin (A + B) \cr}
\]
\[ \eqalign{
& \sin A + \sin B = 2\sin {1 \over 2}(A + B)\cos {1 \over 2}(A - B) \cr
& \sin A - \sin B = 2\cos {1 \over 2}(A + B)\sin {1 \over 2}(A - B) \cr
& \cos A + \cos B = 2\cos {1 \over 2}(A + B)\cos {1 \over 2}(A - B) \cr
& \cos A - \cos B = - 2\sin {1 \over 2}(A + B)\sin {1 \over 2}(A - B) \cr}
\]
1
Thomas' Calculus Early Transcendentals - George B. Thomas Jr., Maurice D. Weir, Joel R. Hass - 13th Edition
نویسنده
علیرضا گلمکانی
شماره کلید
1600
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