FRAC{\sin 3B}{\sin B}=\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=3-4{{\sin }^{2}}B\] \[=3-4+4{{\cos }^{2}}B=-1+\frac{4{{({{a}^{2}}+{{C}^{2}}-{{b}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=-1+\frac{{{\left( \frac{{{a}^{2}}+{{c}^{2}}}{2} \RIGHT)}^{2}}}{{{(ac)}^{2}}}=-1+\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}-4{{a}^{2}}{{c}^{2}}}{4{{(ac)}^{2}}}={{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ac} \right)}^{2}}\].

"> FRAC{\sin 3B}{\sin B}=\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=3-4{{\sin }^{2}}B\] \[=3-4+4{{\cos }^{2}}B=-1+\frac{4{{({{a}^{2}}+{{C}^{2}}-{{b}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=-1+\frac{{{\left( \frac{{{a}^{2}}+{{c}^{2}}}{2} \RIGHT)}^{2}}}{{{(ac)}^{2}}}=-1+\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}-4{{a}^{2}}{{c}^{2}}}{4{{(ac)}^{2}}}={{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ac} \right)}^{2}}\].

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If in \[\Delta \,ABC\], \[2{{b}^{2}}={{a}^{2}}+{{c}^{2}},\]then \[\frac{\sin 3B}{\sin B}=\] [UPSEAT 1999]

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) 10 months ago

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\[\FRAC{\sin 3B}{\sin B}=\frac{3\sin B-4{{\sin }^{3}}B}{\sin B}=3-4{{\sin }^{2}}B\] \[=3-4+4{{\cos }^{2}}B=-1+\frac{4{{({{a}^{2}}+{{C}^{2}}-{{b}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=-1+\frac{{{\left( \frac{{{a}^{2}}+{{c}^{2}}}{2} \RIGHT)}^{2}}}{{{(ac)}^{2}}}=-1+\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}}{4{{(ac)}^{2}}}\] \[=\frac{{{({{a}^{2}}+{{c}^{2}})}^{2}}-4{{a}^{2}}{{c}^{2}}}{4{{(ac)}^{2}}}={{\left( \frac{{{c}^{2}}-{{a}^{2}}}{2ac} \right)}^{2}}\].

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Posted on 18 Sep 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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