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UniLink"}],["$","meta","3",{"name":"description","content":"△ABCにおいて、残りの辺の長さと角の大きさを求める問題（a=2√3、B=15°、C=45°）A=120°とc=2√2まで出せました。残りのbは余弦定理で出すんだと思うのですが、cosCを使うと何度してもb=√6±√2になります。解答はcosAを使ってb=−√2±√6になってます。解答の意味は理解できるのですが、なぜ同じbを求めるのにcosAの余弦定理だとbが違う答えが出てしまうのでしょうか。このやり方でも答えが出る方、途中式込みの解答をどうか教えてください。。"}],["$","link","4",{"rel":"icon","href":"/favicon.ico","type":"image/x-icon","sizes":"48x48"}],["$","link","5",{"rel":"icon","href":"/icon.png?2c0dc65a59843333","type":"image/png","sizes":"180x180"}],["$","link","6",{"rel":"apple-touch-icon","href":"/apple-icon.png?2c0dc65a59843333","type":"image/png","sizes":"180x180"}],["$","meta","7",{"name":"next-size-adjust"}]] 1:null 13:I[3903,["51","static/chunks/795d4814-03346c8d233b4adb.js","212","static/chunks/212-70508e17017a12c2.js","231","static/chunks/231-5dc9f3acdba63b0c.js","54","static/chunks/54-f848f8ba1c362ca7.js","23","static/chunks/app/advice/%5Bid%5D/page-53a773e0095d4429.js"],"ClientInfo"] 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1a:I[6549,["51","static/chunks/795d4814-03346c8d233b4adb.js","212","static/chunks/212-70508e17017a12c2.js","231","static/chunks/231-5dc9f3acdba63b0c.js","54","static/chunks/54-f848f8ba1c362ca7.js","23","static/chunks/app/advice/%5Bid%5D/page-53a773e0095d4429.js"],"CommentItemName"] 1d:I[3866,["51","static/chunks/795d4814-03346c8d233b4adb.js","212","static/chunks/212-70508e17017a12c2.js","231","static/chunks/231-5dc9f3acdba63b0c.js","54","static/chunks/54-f848f8ba1c362ca7.js","23","static/chunks/app/advice/%5Bid%5D/page-53a773e0095d4429.js"],"AdOnAdviceList1"] 1b:T1307,　確かに解法暗記は大切です。しかし、それを単純暗記で終わらせてしまっては危険です。京大の整数問題を例に見ていきましょう。 「n^3ー7n＋9が素数となるような整数nを全て求めよ。」（2018） 　この問題は、整数kを用いて、nを3k、3k＋1、3kー1とに場合分けして考えればすぐ解けます。しかし、この解法を単純に暗記しても、どこからこの解法を導く着想を得たのかが分からなければ、同じ解法を使う問題に対峙してもそれを見抜くことは困難です。この問題では、n＝1を仮に入れてみると、値は3で素数です。次に、n＝2を入れてみた場合、こちらも値は3で素数です。n＝3の場合は15で素数ではない、n＝4の場合は45で素数ではない、n＝5の場合は99で素数ではない……。ここで何か気づくでしょう。すなわち、実験して得られた値は全て3の倍数になっていることに気づくはずです。となれば、与式の取りうる値は全部3の倍数なんじゃないか？という疑いが生じるでしょう。この仮説を確かめるために、まずはすべてのnに対し与式の値は必ず3の倍数になるということを証明すればよいことになり、そのためにnを3で割った余りに注目して場合分けをするという解法に辿り着くわけです（したがって、modを使えばもっと楽な計算で証明できます）。(i)n＝3kの場合は言うまでもないとして、(ii)n＝3k＋1の場合、与式は27k^3＋27k^2ー12k＋3で、(iii)n＝3kー1の場合、27k^3ー27k^2ー12k＋15で、いずれも3の倍数になります。素数の中で3の倍数は3だけなので、結局この問題は、（与式）＝3という方程式を整数nについて解けば良いということになります。 　 　こんな感じで解法を深く見つめていくと、解ける問題も増えていきます。例えば、この問題。 「pが素数ならばp^4＋14は素数でないことを示せ。」（2021文系） 　p＝2のとき値は30、p＝3のとき値は95、p＝5のとき値は639、p＝7のとき値は2415、p＝11のとき値は14655……。p＝3のとき以外は、いずれも3の倍数です。よって、(i)p＝3のときと、(ii)p ≠ 3の時で場合分けをして、(ii)p ≠ 3のときでは、さらに(a)p ≡ 1（mod3）のときと、(b)p ≡ 2（mod3）のときとで場合分けして、p^4＋14が素数pに対し常に3の倍数となることを証明し、そのとき取りうる値は3のみであるが、p^4＋14はp＝2で最小値30であるから、3を取ることはない。したがって、p^4＋14は素数ではない、という解決ができるわけです。 　 　また、この問題も。 「素数p, qを用いて、p^q＋q^pと表される素数をすべて求めよ。」（2016理系） 　pとqの対称性からp≦qとしても一般性は失われないので、この大小関係のもと進めていきます。まず、2数の偶奇が一致するとき、その和は必ず偶数になりますが、pとqはいずれも素数なので、与式の取りうる値は最小でも8（p＝2, q＝2）であり、値が2となることはありません。このことから、与式の値は奇数であり、そのためにはp＝2でなければなりません（片方は偶数でなければならず、p^qが偶数となるのはp＝2の場合だけ）。すると、p＝2と固定して、qに3、5、7、11……と入れてみればいいわけです。q＝3のとき値は17で素数、q＝5のとき値は57で素数ではない、q＝7のとき値は177で素数ではない、q＝11のとき値は2169で素数ではない……。q＝3のときを除いて、すべて3の倍数ですね。しかし、この問題では、安易にqを3で割った余りで場合分けしてもうまくいきません。場合分けにさらなる工夫が必要になりますが、そこは自力でやってみましょう。 　上の問題は、いずれも同じところから解法の着想を得ていることがわかったと思います。と同時に、個別の問題にだけ通用するような覚え方をしても、似た問題ですら手が止まってしまうということも。やはり何事も、勉強というからには自分の頭で考えなければなりません。ただ単に、与えられた結果の知識や表現を覚えるだけではダメですね。その点、受験勉強は大変なものですが、そういったことも志望校という目標に向かって一途に続けられる人こそ、本番で勝っていく人たちなのでしょう。私も偉そうなことは言えませんがね。1c:T13f3,こんにちは！RIZと申します。 問題集の問題は解けるけれど初見の問題では解けなくなるということですね。 まずとても当たり前の話をしますが、数学は問題文から解答を考えなければなりません。現在の、問題集の問題は解けるけれども初見の問題では手が止まってしまうというのは、単に問題集の答えを覚えているだけに他なりません。そこで、今回は初見の問題でも解けるようにするためにはどのようにすれば良いかについてお話しします。 前提として、数学の公式や定義はしっかり学習しているとします。もし質問文に書かれている数学用語というのがこうした公式や定義であるなら、定義はまずしっかり覚えてください。そして公式についてはできれば丸暗記するより、導出できるようにしたほうが良いです。ただもう時間があまりないので最悪丸暗記でもいいですが、導出できるようにすることで、なぜその公式が成り立つのか理解できるので覚えやすくもなりますし、もし忘れてしまっても対応できるようになるのでおすすめです。例えば三角関数の2倍角とか3倍角なんかは加法定理とか、数3ですがド・モアブルの定理などから簡単に導出できますよね。加法定理を毎回導出するのは流石に面倒ですが、2倍角や3倍角を加法定理から導出するのは少しの時間でできますよね。このようにあまり覚えていなくても簡単に導出できる公式はなるべく導出できるようにした方が良いです。 さて、話を戻しますが、以上のように公式や定義が頭に入っていることを前提として、初見の問題でどのように対処するべきかについてお話しします。まず冒頭でもお話ししたように、数学は問題文だけから解答を考えなければなりません。そこでまず、問題文の条件に着目します。条件というのはいろいろあります。例えばnを自然数とするとか、x、yが円の方程式を満たしているとか、垂直に交わるとか、さまざまです。他にも、直接的には書かれていないけれども重要な条件もあります。例えば与えられた式が対称式であるとかです。こうした条件から、解答を考えていきます。例えば上の例で言えば、nを自然数として、かつnに関する命題が与えられて証明しなさいといった問題であれば、自然数かつ証明問題であることから数学的帰納法が浮かびますし、x、yが円の方程式を満たしていて、かつx、yの2変数からなる関数の最大最小を考えたい時、xとyが円の方程式を満たすという条件から、θを媒介変数としてx、yをcosθとsinθで置くとかが考えられます。他にも、垂直に交わるという条件があれば、例えばその垂直に交わる直線の傾き同士の積は−1とか、内積0とか、あるいは図形的に三平方の定理を利用することも可能かもしれません。以上のように、条件を見たときにいろいろなことが考えられるようになることで、初見の問題で同じような条件が出てきたときに対応できます。もちろん入試問題というのは問題集には載っていない初見の問題である場合がほとんどです。なので普段解いている問題と全く同じでないのは当たり前ですが、条件に関して言えば部分的に共通していますよね。なのでこうしたことが想起できるようになれば、初見の問題でも対応できるようになるわけです。しかしこのように、条件を見てそこから解法を想起するというのは初見では無理ですよね。それを問題集から学ぶわけです。つまり、ただ問題を解いて、解けなかったら答えを見て覚えて終わりではなく、解法を見たとき、それが「なぜ」そうなるのかを考えます。そして、もし自分が初見でその問題を解くとしたら、まず問題文のどの条件に着目するのかを考えます。このようにすることで、解法のストックを増やしていくわけです。とにかく、解答を見たものでも初見だったらどうするのか、そして「なぜ」そうするのかまで説明できるようになることで、初見の問題でも、それまでストックした解法の引き出しから解法を想起でき、対応できるようになるわけです。なのでまずは今までやった問題集で、問題文のどの条件に着目して、「なぜ」その解答になるのか考えながら学習するようにしてみてください。以上になります。ご質問などありましたらコメント欄の方でお願いします！1e:Tf02,センター試験の集合は、実数の集合を扱うことが多いため、数直線上に図示するのが有効なことが多いです。 目盛の間隔を正確に図示する必要はなく、それぞれの端の大小と、黒丸白丸があっているかが重要です。(黒丸の場合はその点を含む、白丸の時はその点を含まないことを表します。不等号に=が入っているかどうかの違いとも言えます。) 例えば、 p: x>1 q:x≦2 のように与えられていた時、右向きの数直線上に左から1と2の点を書きます。 pについては、x>1(つまり「xは1より大きい」)であることから、先ほど書いた1の点に白丸を書き、そこから右上がりに少し直線を書き、そこから右向きに直線を伸ばします。新幹線のような形になります。この形は、1の点を含まないことを表すもので、白丸と同じ意味ですが、ぱっと見で分かるように両方使います。また、この線がpであることをどこかに書いておいてください。 qについては、x≦2(つまり｢xは2以下｣)であるので、2の点に黒丸を書き、そこから真下に少し直線を書き、左向きの直線を伸ばします。こちらは、電車のような形になります。この形は、2を含むことを表すもので、黒丸と同じ意味です。こちらの線にも、qであることを書いておいてください。 このように、範囲を一つ一つ図示していくと、次のようになります。 _______________ p / 2 ---------○-----●------->x 1 | q --------------- これを見れば、「pかつq」や、「pまたはq」「p⇒q は真か偽か」はすぐに分かるはずです。たとえば｢pかつq｣なら、pとqが重なっているところなので、10)が答えとして出るのに\ncosCの場合はb=√6±√2がb>0の条件でどちらも有効で、cosAの時と同じにならないのではないかという疑問だと思って回答しますね！\n\nこの場合cosCで出てきたbの値に対して一つ有効な条件設定があります。それが「辺と角の関係」です。\nもしかしたらこの時点でピンと来たかもしれませんが、角度が大きい角の対面の辺が長くなるよって感じのやつですね(言葉がラフでごめんなさい。回答に書くときはしっかり教科書通りのやつ書いてね笑)\nその関係性を角Cと角Bに当てはめてみると角度が小さい方の対面の辺bは辺cより小さい必要があります。\n√6+√2と2√2の大小関係、√6-√2と2√2の大小関係はどういう風に考えるといいんでしたっけ？\n一回考えてみてください🙆‍♂️(逆にいうとその考えがめんどくさくて回答はcosAを採用したのかもしれませんね…)\n\nまた、他にも考え方があると思うのでこういう考え方もあるよ！ってのを思いついたら是非教えてくださいね🥸\n頑張って！"]}]]}],["$","div",null,{"className":"mb-4","children":["$","$L16",null,{"adviserImageUrl":"https://firebasestorage.googleapis.com/v0/b/unilink-48e75.appspot.com/o/images%2Fs_Oou0tJz4wTbzaJ6HpZzQAgIh9GZ2.jpg?alt=media&token=690b2ef6-c85b-45da-9920-7d340e4fc684","adviserName":"yuya","adviserDepartment":"東京工業大学物質理工学院","adviceId":"anA1oy6OOsttb4sFEbjB","numberOfFan":143,"clipsAvg":11.77319587628866,"adviceRateAvg":4.7554945054945055,"profile":"【経歴】\n公立中学→私立滑り止め高校(都立落ち)→現役東工大→東工大大学院→来年度就職\n\n「受験期に無理な勉強やストレスで何度も体調を崩しました。自分のような人を減らせるように受験生の力になりたいです。」\n\n「ファン」→「メッセージ」で相談乗ります❗️\n連絡ください🙆‍♂️\n※現在指導は募集していません"}]}],["$","div",null,{"children":["$","$L7",null,{"href":"https://ck.jp.ap.valuecommerce.com/servlet/referral?sid=3364577&pid=884970531&vc_url=http%3A%2F%2Fshingakunet.com%2F%3Fvos%3Dnrmnvccp0000100","rel":"nofollow","target":"_blank","children":["$","$L8",null,{"src":"/images/document_request_banner.jpg","width":3660,"height":1500,"sizes":"100vw","style":{"width":"100%","height":"auto"},"alt":"UniLink 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mb-1","children":"質問者様は高2ということなので、数Ⅱまでの範囲で回答させていただきます。\n\n\n【三角関数を変形する目的】\n\nまず、三角関数を変形するのは必ず目的があります。\n①三角関数を含んだ方程式・不等式を解くため\n②三角関数を含んだ関数の最大値・最小値を求めるため\nなどがよくある目的ですね。\n\n《①について》\n方程式や不等式ははじめに因数分解で攻めます。\n(因数)(因数)=0\nといった形になれば、あとは簡単ですね。\n因数分解しない場合は②の考え方をそのまま借りましょう\n\n《②について》\nsinのみ、cosのみ、tanのみ、の式に帰着させます。そしたら見たことある関数(一次関数、二次関数など)になります。\nそのための手段として\n＊三角関数の相互関係\n＊加法定理を用いた公式\nなどが存在します。\n\n\n－－－－－－－－－\n\n【質問主様の弱点と思われるところ】\n\n数Ⅱの三角関数に入ってからうまくいかなくなった高校生は加法定理を用いた公式につまづいている人が多いです。\n公式自体覚えていても、問題でうまく活用出来ないことがよくあります。\n\n先程の項目で書きました、変形のそもそもの目的を意識して演習してみてください。\n使い分けパターンは青チャートなどのテキストに詳しく記載されています。これを身につけることが大切です。\n\nパターンを繰り返しの演習で身につける際に、\n「因数分解を目指す！」\n「sinのみ、cosのみ、tanのみの式を目指す！」\nという意識を持って取り組むことで、何故その式変形を使うのかが体感出来ます。\n\n\n－－－－－－－－－\n\n【最後に】\n\n問題のゴールから逆算して考えることが数学においては大切です。\n初めから逆算して考えることなんて出来ないから、パターンを演習によって身につけるわけですが、ゴールを意識してパターンを身につけなければ、何のためのパターンなのかがわかりません。\n必ず、式変形の目的を意識した演習を心掛けてください。"}],["$","div",null,{"className":"flex 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